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21673
|
\section{Sphere is Set Difference of Closed Ball with Open Ball/Normed Division Ring}
Tags: Normed Division Rings, Normed Division Ring
\begin{theorem}
Let $\struct{R, \norm {\,\cdot\,} }$ be a normed division ring.
Let $a \in R$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-closed ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {B_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-open ball of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Let $\map {S_\epsilon} {a; \norm {\,\cdot\,} }$ denote the $\epsilon$-sphere of $a$ in $\struct {R, \norm {\,\cdot\,} }$.
Then:
:$\map {S_\epsilon} {a; \norm {\,\cdot\,} } = \map { {B_\epsilon}^-} {a; \norm {\,\cdot\,} } \setminus \map {B_\epsilon} {a; \norm {\,\cdot\,} }$
\end{theorem}
\begin{proof}
The result follows directly from:
:Closed Ball in Normed Division Ring is Closed Ball in Induced Metric
:Open Ball in Normed Division Ring is Open Ball in Induced Metric
:Sphere in Normed Division Ring is Sphere in Induced Metric
:Sphere is Set Difference of Closed Ball with Open Ball
{{qed}}
Category:Normed Division Rings
\end{proof}
|
21674
|
\section{Sphere is Set Difference of Closed Ball with Open Ball/P-adic Numbers}
Tags: P-adic Number Theory
\begin{theorem}
Let $p$ be a prime number.
Let $\Q_p$ be the $p$-adic numbers.
Let $a \in \Q_p$.
Let $\epsilon \in \R_{>0}$ be a strictly positive real number.
Let $\map {{B_\epsilon}^-} a$ denote the $\epsilon$-closed ball of $a$ in $\Q_p$.
Let $\map {B_\epsilon} a$ denote the $\epsilon$-open ball of $a$ in $\Q_p$.
Let $\map {S_\epsilon} a$ denote the $\epsilon$-sphere of $a$ in $\Q_p$.
Then:
:$\map {S_\epsilon} a = \map { {B_\epsilon}^-} a \setminus \map {B_\epsilon} a$
\end{theorem}
\begin{proof}
The result follows directly from:
:P-adic Closed Ball is Instance of Closed Ball of a Norm
:P-adic Open Ball is Instance of Open Ball of a Norm
:P-adic Sphere is Instance of Sphere of a Norm
:Sphere is Set Difference of Closed and Open Ball in Normed Division Ring
{{qed}}
Category:P-adic Number Theory
\end{proof}
|
21675
|
\section{Spherical Law of Cosines}
Tags: Spherical Law of Cosines, Spherical Trigonometry, Named Theorems
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos a = \cos b \cos c + \sin b \sin c \cos A$
\end{theorem}
\begin{proof}
:500px
Let $A$, $B$ and $C$ be the vertices of a spherical triangle on the surface of a sphere $S$.
By definition of a spherical triangle, $AB$, $BC$ and $AC$ are arcs of great circles on $S$.
By definition of a great circle, the center of each of these great circles is $O$.
Let $AD$ be the tangent to the great circle $AB$.
Let $AE$ be the tangent to the great circle $AC$.
Thus the radius $OA$ of $S$ is perpendicular to $AD$ and $AE$.
By construction, $AD$ lies in the same plane as $AB$.
Thus when $OB$ is produced, it will intersect $AD$ at $D$, say.
Similarly, $OC$ can be produced to intersect $AE$ at $E$, say.
The spherical angle $\sphericalangle BAC$ is defined as the angle between the tangents $AD$ and $AE$.
Thus:
:$\sphericalangle BAC = \angle DAE$
or, denoting that spherical angle $\sphericalangle BAC$ as $A$:
:$A = \angle DAE$
In the (plane) triangle $OAD$, we have that $\angle OAD$ is a right angle.
We also have that $\angle AOD = \angle AOB$ is equal to $c$, by definition of the length of a side of a spherical triangle.
Thus:
{{begin-eqn}}
{{eqn | l = AD
| r = OA \tan c
| c =
}}
{{eqn | l = OD
| r = OA \sec c
| c =
}}
{{end-eqn}}
and by similar analysis of $\triangle OAE$, we have:
{{begin-eqn}}
{{eqn | l = AE
| r = OA \tan b
| c =
}}
{{eqn | l = OE
| r = OA \sec b
| c =
}}
{{end-eqn}}
From consideration of $\triangle DAE$:
{{begin-eqn}}
{{eqn | l = DE^2
| r = AD^2 + AE^2 - 2 AD \cdot AE \cos \angle DAE
| c = Law of Cosines
}}
{{eqn | n = 1
| r = OA^2 \paren {\tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A}
| c =
}}
{{end-eqn}}
From consideration of $\triangle DOE$:
{{begin-eqn}}
{{eqn | l = DE^2
| r = OD^2 + OE^2 - 2 OD \cdot OE \cos \angle DOE
| c = Law of Cosines
}}
{{eqn | n = 2
| r = OA^2 \paren {\sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a}
| c = as $\angle DOE = \angle BOC$
}}
{{end-eqn}}
Thus:
{{begin-eqn}}
{{eqn | l = \sec^2 c + \sec^2 b - 2 \sec b \sec c \cos a
| r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A
| c = from $(1)$ and $(2)$
}}
{{eqn | ll= \leadsto
| l = \paren {1 + \tan^2 c} + \paren {1 + \tan^2 b} - 2 \sec b \sec c \cos a
| r = \tan^2 c + \tan^2 b - 2 \tan b \tan c \cos A
| c = Difference of Squares of Secant and Tangent
}}
{{eqn | ll= \leadsto
| l = 1 - \sec b \sec c \cos a
| r = \tan b \tan c \cos A
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \cos b \cos c - \cos a
| r = \sin b \sin c \cos A
| c = multiplying both sides by $\cos b \cos c$
}}
{{end-eqn}}
and the result follows.
{{qed}}
\end{proof}
|
21676
|
\section{Spherical Law of Cosines/Angles}
Tags: Spherical Trigonometry, Named Theorems
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\cos A = -\cos B \cos C + \sin B \sin C \cos a$
\end{theorem}
\begin{proof}
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Let the sides $a', b', c'$ of $\triangle A'B'C'$ be opposite $A', B', C'$ respectively.
From Spherical Triangle is Polar Triangle of its Polar Triangle we have that:
:not only is $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$
:but also $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
We have:
{{begin-eqn}}
{{eqn | l = \cos a'
| r = \cos b' \cos c' + \sin b' \sin c' \cos A'
| c = Spherical Law of Cosines
}}
{{eqn | ll= \leadsto
| l = \map \cos {\pi - A}
| r = \map \cos {\pi - B} \, \map \cos {\pi - C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \map \cos {\pi - a}
| c = Side of Spherical Triangle is Supplement of Angle of Polar Triangle
}}
{{eqn | ll= \leadsto
| l = -\cos A
| r = \paren {-\cos B} \paren {-\cos C} + \map \sin {\pi - B} \, \map \sin {\pi - C} \, \paren {-\cos a}
| c = Cosine of Supplementary Angle
}}
{{eqn | ll= \leadsto
| l = -\cos A
| r = \paren {-\cos B} \paren {-\cos C} + \sin B \sin C \paren {-\cos a}
| c = Sine of Supplementary Angle
}}
{{eqn | ll= \leadsto
| l = \cos A
| r = -\cos B \cos C + \sin B \sin C \cos a
| c = simplifying and rearranging
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21677
|
\section{Spherical Law of Sines}
Tags: Spherical Law of Sines, Spherical Trigonometry, Named Theorems
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sin b \sin c \cos A
| r = \cos a - \cos b \cos c
| c = Spherical Law of Cosines
}}
{{eqn | ll= \leadsto
| l = \sin^2 b \sin^2 c \cos^2 A
| r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
| c =
}}
{{eqn | ll= \leadsto
| l = \sin^2 b \sin^2 c \paren {1 - \sin^2 A}
| r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
| l = \sin^2 b \sin^2 c - \sin^2 b \sin^2 c \sin^2 A
| r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
| c = multiplying out
}}
{{eqn | ll= \leadsto
| l = \paren {1 - \cos^2 b} \paren {1 - \cos^2 c} - \sin^2 b \sin^2 c \sin^2 A
| r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
| c = Sum of Squares of Sine and Cosine
}}
{{eqn | ll= \leadsto
| l = 1 - \cos^2 b - \cos^2 c + \cos^2 b \cos^2 c - \sin^2 b \sin^2 c \sin^2 A
| r = \cos^2 a - 2 \cos a \cos b \cos c + \cos^2 b \cos^2 c
| c = multiplying out
}}
{{eqn | n = 1
| ll= \leadsto
| l = \sin^2 b \sin^2 c \sin^2 A
| r = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c
| c = rearranging and simplifying
}}
{{end-eqn}}
Let $X \in \R_{>0}$ such that:
:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
Then from $(1)$:
{{begin-eqn}}
{{eqn | l = \dfrac {X^2 \sin^2 a \sin^2 b \sin^2 c} {\sin^2 b \sin^2 c \sin^2 A}
| o = =
| r = \dfrac {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c} {1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c}
| c =
}}
{{eqn | ll= \leadsto
| l = X^2
| r = \dfrac {\sin^2 A} {\sin^2 a}
| c =
}}
{{end-eqn}}
In a spherical triangle, all of the sides are less than $\pi$ radians.
The same applies to the angles.
From Shape of Sine Function:
:$\sin \theta > 0$ for all $0 < \theta < \pi$
Hence the negative root of $\dfrac {\sin^2 A} {\sin^2 a}$ does not apply, and so:
:$X = \dfrac {\sin A} {\sin a}$
Similarly, from applying the Spherical Law of Cosines to $\cos B$ and $\cos C$:
{{begin-eqn}}
{{eqn | l = \sin a \sin c \cos B
| r = \cos b - \cos a \cos c
}}
{{eqn | l = \sin a \sin b \cos C
| r = \cos c - \cos a \cos b
}}
{{end-eqn}}
we arrive at the same point:
{{begin-eqn}}
{{eqn | l = X
| r = \dfrac {\sin B} {\sin b}
}}
{{eqn | r = \dfrac {\sin A} {\sin a}
}}
{{end-eqn}}
where:
:$X^2 \sin^2 a \sin^2 b \sin^2 c = 1 - \cos^2 a - \cos^2 b - \cos^2 c + 2 \cos a \cos b \cos c$
as before.
Hence we have:
:$\dfrac {\sin a} {\sin A} = \dfrac {\sin b} {\sin B} = \dfrac {\sin c} {\sin C}$
{{qed}}
\end{proof}
|
21678
|
\section{Spherical Law of Tangents}
Tags: Spherical Trigonometry, Named Theorems
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Then:
:$\dfrac {\tan \frac 1 2 \paren {A + B} } {\tan \frac 1 2 \paren {A - B} } = \dfrac {\tan \frac 1 2 \paren {a + b} } {\tan \frac 1 2 \paren {a - b} }$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \tan \dfrac {A + B} 2
| r = \dfrac {\cos \frac {a - b} 2} {\cos \frac {a + b} 2} \cot \dfrac C 2
| c = Napier's Analogies
}}
{{eqn | n = 1
| ll= \leadsto
| l = \tan \frac {A + B} 2 \cos \frac {a + b} 2
| r = \cos \frac {a - b} 2 \cot \frac C 2
| c = more manageable in this form
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = \tan \dfrac {A - B} 2
| r = \dfrac {\sin \frac {a - b} 2} {\sin \frac {a + b} 2} \cot \dfrac C 2
| c = Napier's Analogies
}}
{{eqn | n = 2
| ll= \leadsto
| l = \tan \frac {A - B} 2 \sin \frac {a + b} 2
| r = \sin \frac {a - b} 2 \cot \frac C 2
| c = more manageable in this form
}}
{{end-eqn}}
Hence we have:
{{begin-eqn}}
{{eqn | l = \dfrac {\tan \frac {A + B} 2 \cos \frac {a + b} 2} {\tan \frac {A - B} 2 \sin \frac {a + b} 2}
| r = \dfrac {\cos \frac {a - b} 2 \cot \frac C 2} {\sin \frac {a - b} 2 \cot \frac C 2}
| c = dividing $(1)$ by $(2)$
}}
{{eqn | ll= \leadsto
| l = \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2} \dfrac 1 {\tan \frac {a + b} 2}
| r = \dfrac 1 {\tan \frac {a - b} 2}
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \dfrac {\tan \frac {A + B} 2} {\tan \frac {A - B} 2}
| r = \dfrac {\tan \frac {a + b} 2} {\tan \frac {a - b} 2}
| c = simplifying
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21679
|
\section{Spherical Triangle is Polar Triangle of its Polar Triangle}
Tags: Polar Triangles
\begin{theorem}
Let $\triangle ABC$ be a spherical triangle on the surface of a sphere whose center is $O$.
Let the sides $a, b, c$ of $\triangle ABC$ be measured by the angles subtended at $O$, where $a, b, c$ are opposite $A, B, C$ respectively.
Let $\triangle A'B'C'$ be the polar triangle of $\triangle ABC$.
Then $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
\end{theorem}
\begin{proof}
:400px
Let $BC$ be produced to meet $A'B'$ and $A'C'$ at $L$ and $M$ respectively.
Because $A'$ is the pole of the great circle $LBCM$, the spherical angle $A'$ equals the side of the spherical triangle $ALM$.
By construction we have that $B'$ is the pole of $AC$.
Thus the length of the arc of the great circle from $B$ to any point on $AC$ is a right angle.
Similarly, the length of the arc of the great circle from $A'$ to any point on $BC$ is also a right angle.
Hence:
:the length of the great circle arc $CA'$ is a right angle
:the length of the great circle arc $CB'$ is a right angle
and it follows by definition that $C$ is a pole of $A'B'$.
In the same way:
:$A$ is a pole of $B'C'$
:$B$ is a pole of $A'C'$.
Hence, by definition, $\triangle ABC$ is the polar triangle of $\triangle A'B'C'$.
{{qed}}
\end{proof}
|
21680
|
\section{Split Epimorphism is Epic}
Tags: Category Theory
\begin{theorem}
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a split epimorphism.
Then $f: C \twoheadrightarrow D$ is epic.
\end{theorem}
\begin{proof}
Let $g: D \to C$ be a morphism such that:
:$f \circ g = \operatorname{id}_D$
which is guaranteed to exist by definition of split epimorphism.
Suppose that $x, y: D \to E$ are morphisms such that:
:$x \circ f = y \circ f$
Then necessarily also:
:$x \circ f \circ g = y \circ f \circ g$
and hence, since $f \circ g = \operatorname{id}_D$, it follows that:
:$x \circ \operatorname{id}_D = y \circ \operatorname{id}_D$
which yields the result by the definition of identity morphism.
The situation is illustrated by the following commutative diagram:
::$\begin{xy}
<0em,0em> *+{C} = "C",
<0em,-4em>*+{D} = "D",
<4em,0em> *+{D} = "D2",
<8em,0em> *+{E} = "E",
"D2"+/r.5em/+/^.25em/;"E"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x},
"D2"+/r.5em/+/_.25em/;"E"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y},
"C";"D2" **@{-} ?>*@{>} ?*!/_.6em/{f},
"D";"D2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C},
"D";"C" **@{-} ?>*@{>} ?*!/_.6em/{g},
\end{xy}$
{{qed}}
\end{proof}
|
21681
|
\section{Split Monomorphism is Monic}
Tags: Category Theory
\begin{theorem}
Let $\mathbf C$ be a metacategory.
Let $f: C \to D$ be a split monomorphism.
Then $f: C \rightarrowtail D$ is monic.
\end{theorem}
\begin{proof}
Let $g: D \to C$ be a morphism such that:
:$g \circ f = \operatorname{id}_C$
which is guaranteed to exist by definition of split monomorphism.
Suppose that $x, y: B \to C$ are morphisms such that:
:$f \circ x = f \circ y$
Then necessarily also:
:$g \circ f \circ x = g \circ f \circ y$
and hence, since $g \circ f = \operatorname{id}_C$, it follows that:
:$\operatorname{id}_C \circ x = \operatorname{id}_C \circ y$
which yields the result by the definition of identity morphism.
The situation is illustrated by the following commutative diagram:
::$\begin{xy}
<-4em,0em>*+{B} = "B",
<0em,0em> *+{C} = "C",
<4em,0em> *+{D} = "D",
<4em,-4em>*+{C} = "C2",
"B"+/r.5em/+/^.25em/;"C"+/l.5em/+/^.25em/ **@{-} ?>*@{>} ?*!/_.6em/{x},
"B"+/r.5em/+/_.25em/;"C"+/l.5em/+/_.25em/ **@{-} ?>*@{>} ?*!/^.6em/{y},
"C";"D" **@{-} ?>*@{>} ?*!/_.6em/{f},
"C";"C2" **@{-} ?>*@{>} ?*!/^.6em/{\operatorname{id}_C},
"D";"C2" **@{-} ?>*@{>} ?*!/_.6em/{g},
\end{xy}$
{{qed}}
\end{proof}
|
21682
|
\section{Sprague's Property of Root 2}
Tags: Square Root of 2, Beatty Sequences
\begin{theorem}
Let $S = \sequence {s_n}$ be the sequence of fractions defined as follows:
Let the numerator of $s_n$ be:
:$\floor {n \sqrt 2}$
where $\floor x$ denotes the floor of $x$.
Let the denominators of the terms of $S$ be the (strictly) positive integers missing from the numerators of $S$:
:$S := \dfrac 1 3, \dfrac 2 6, \dfrac 4 {10}, \dfrac 5 {13}, \dfrac 7 {17}, \dfrac 8 {20}, \ldots$
Then the difference between the numerator and denominator of $s_n$ is equal to $2 n$.
\end{theorem}
\begin{proof}
Denote the numerators of the terms of $S$ as $\sequence {N_n}$.
Denote the denominators of the terms of $S$ as $\sequence {D_n}$.
From the definition:
:$\sequence {N_n}$ is a Beatty sequence, where $\sequence {N_n} = \BB_{\sqrt 2} = \sequence{\floor{n \sqrt 2} }_{n \mathop \in \Z_{> 0} }$
:$\sequence {N_n}$ and $\sequence {D_n}$ are complementary Beatty sequences.
Then by Beatty's Theorem, $\sequence {D_n}$ is a Beatty sequence.
Define $\sequence {D_n} = \BB_y = \sequence{\floor{n y} }_{n \mathop \in \Z_{> 0} }$.
Then we have:
{{begin-eqn}}
{{eqn | l = \dfrac 1 {\sqrt 2} + \dfrac 1 y
| r = 1
| c = Beatty's Theorem
}}
{{eqn | ll = \leadsto
| l = \dfrac 1 y
| r = 1 - \dfrac 1 {\sqrt 2}
}}
{{eqn | r = \dfrac {\sqrt 2 - 1} {\sqrt 2}
}}
{{eqn | ll = \leadsto
| l = y
| r = \dfrac {\sqrt 2} {\sqrt 2 - 1}
}}
{{eqn | r = \dfrac {\sqrt 2 \paren {\sqrt 2 + 1} } {\paren {\sqrt 2 - 1} \paren {\sqrt 2 + 1} }
}}
{{eqn | r = 2 + \sqrt 2
}}
{{end-eqn}}
So we have $s_n = \dfrac {N_n} {D_n} = \dfrac {\floor {n \sqrt 2} } {\floor {n \paren {2 + \sqrt 2} } }$.
The difference between the numerator and denominator of $s_n$ is $\floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}$.
We have:
{{begin-eqn}}
{{eqn | l = \floor {n \paren {2 + \sqrt 2} } - \floor {n \sqrt 2}
| r = \floor {2 n + n \sqrt 2} - \floor {n \sqrt 2}
}}
{{eqn | r = 2 n + \floor {n \sqrt 2} - \floor {n \sqrt 2}
| c = Floor of Number plus Integer
}}
{{eqn | r = 2 n
}}
{{end-eqn}}
Hence the result.
{{qed}}
{{Namedfor|Roland Percival Sprague|cat = Sprague, Roland}}
\end{proof}
|
21683
|
\section{Square-Summable Indexed Sets Closed Under Addition}
Tags: Generalized Sums
\begin{theorem}
Let $\family {a_i}_{i \mathop \in I}, \family {b_i}_{i \mathop \in I}$ be $I$-indexed families of real numbers.
Let:
:$\ds \sum \set {a_i^2: i \in I} < \infty$
:$\ds \sum \set {b_i^2: i \in I} < \infty$
where $\ds \sum$ denotes the generalized sums.
Then:
:$\ds \sum \set {\paren {a_i + b_i}^2: i \in I} < \infty$
\end{theorem}
\begin{proof}
{{proof wanted|Some pain in the behind with careful considerations}}
Category:Generalized Sums
\end{proof}
|
21684
|
\section{Square Cullen Numbers}
Tags: Cullen Numbers, Square Numbers
\begin{theorem}
The numbers:
:$1, 9, 25$
are Cullen numbers which are also square.
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 1
| r = 0 \times 2^0 + 1
}}
{{eqn | l = 9
| r = 2 \times 2^2 + 1
}}
{{eqn | l = 25
| r = 3 \times 2^3 + 1
}}
{{end-eqn}}
{{Qed}}
\end{proof}
|
21685
|
\section{Square Divides Product of Multiples}
Tags: Divisors
\begin{theorem}
Let $a, b, c, \in \Z$ be integers.
Let:
:$a \divides b, a \divides c$
where $\divides$ denotes divisibility.
Then:
:$a^2 \divides b c$
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = a
| o = \divides
| r = b
| c =
}}
{{eqn | ll= \leadsto
| q = \exists k_1 \in \Z
| l = k_1 a
| r = b
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | l = a
| o = \divides
| r = c
| c =
}}
{{eqn | ll= \leadsto
| q = \exists k_2 \in \Z
| l = k_2 a
| r = c
| c = {{Defof|Divisor of Integer}}
}}
{{eqn | ll= \leadsto
| l = b c
| r = \paren {k_1 a} \paren {k_2 a}
| c =
}}
{{eqn | r = \paren {k_1 k_2} a^2
| c =
}}
{{eqn | ll= \leadsto
| l = a^2
| o = \divides
| r = b c
| c = {{Defof|Divisor of Integer}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21686
|
\section{Square Inscribed in Circle is greater than Half Area of Circle}
Tags: Circles, Squares
\begin{theorem}
A square inscribed in a circle has an area greater than half that of the circle.
\end{theorem}
\begin{proof}
:300px
Let $ABCD$ be a square inscribed in a circle.
Let $EFGH$ be a square circumscribed around the same circle.
We have that:
:$ABCD$ is twice the area of the triangle $ADB$.
:$EFGH$ is twice the area of the rectangle $EFBD$.
From Area of Rectangle, the area of $EFBD$ is $ED \cdot DB$.
From Area of Triangle in Terms of Side and Altitude, the area of $ADB$ is $\dfrac 1 2 \cdot ED \cdot DB$.
Thus:
:$EFGH = 2 \cdot ABCD$
But the area of $EFGH$ is greater than the area of the circle around which it is circumscribed.
Therefore half of the area of $EFGH$ is greater than half of the area of the circle around which it is circumscribed.
Therefore the area of $ABCD$ is greater than half of the area of the circle within which it is inscribed.
{{qed}}
\end{proof}
|
21687
|
\section{Square Matrices over Real Numbers under Multiplication form Monoid}
Tags: Monoids, Matrix Algebra, Examples of Monoids
\begin{theorem}
Let $\map {\MM_\R} n$ be a $n \times n$ matrix space over the set of real numbers $\R$.
Then the set of all $n \times n$ real matrices $\map {\MM_\R} n$ under matrix multiplication (conventional) forms a monoid.
\end{theorem}
\begin{proof}
:Matrix Multiplication over Order n Square Matrices is Closed.
:Matrix Multiplication is Associative.
:The Unit Matrix is Unity of Ring of Square Matrices.
{{qed}}
\end{proof}
|
21688
|
\section{Square Matrices with +1 or -1 Determinant under Multiplication forms Group}
Tags: Matrix Groups, Examples of Groups
\begin{theorem}
Let $n \in \Z_{>0}$ be a strictly positive integer.
Let $S$ be the set of square matrices of order $n$ of real numbers whose determinant is either $1$ or $-1$.
Let $\struct {S, \times}$ denote the algebraic structure formed by $S$ whose operation is (conventional) matrix multiplication.
Then $\struct {S, \times}$ is a group.
\end{theorem}
\begin{proof}
Taking the group axioms in turn:
\end{proof}
|
21689
|
\section{Square Matrix is Row Equivalent to Triangular Matrix}
Tags: Matrix Algebra, Triangular Matrices, Square Matrices, Determinants, Square Matrix is Row Equivalent to Triangular Matrix
\begin{theorem}
Let $\mathbf A = \sqbrk a_n$ be a square matrix of order $n$ over a commutative ring $R$.
Then $\mathbf A$ can be converted to an upper or lower triangular matrix by elementary row operations.
\end{theorem}
\begin{proof}
Let $\mathbf A$ be a square matrix of order $n$.
We proceed by induction on $n$, the number of rows of $\mathbf A$.
\end{proof}
|
21690
|
\section{Square Matrix with Duplicate Columns has Zero Determinant}
Tags: Square Matrix with Duplicate Columns has Zero Determinant, Determinants, Matrix Algebra
\begin{theorem}
If two columns of a square matrix over a commutative ring $\struct {R, +, \circ}$ are identical, then its determinant is zero.
\end{theorem}
\begin{proof}
Let $\mathbf A$ be a square matrix over $R$ with two identical columns.
Let $\mathbf A^\intercal$ denote the transpose of $\mathbf A$.
Then $\mathbf A^\intercal$ has two identical rows.
Then:
{{begin-eqn}}
{{eqn | l = \map \det {\mathbf A}
| r = \map \det {\mathbf A^\intercal}
| c = Determinant of Transpose
}}
{{eqn | r = 0
| c = Square Matrix with Duplicate Rows has Zero Determinant
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21691
|
\section{Square Matrix with Duplicate Rows has Zero Determinant}
Tags: Square Matrix with Duplicate Rows has Zero Determinant, Determinants, Matrix Algebra, Matrices
\begin{theorem}
If two rows of a square matrix over a commutative ring $\struct {R, +, \circ}$ are the same, then its determinant is zero.
\end{theorem}
\begin{proof}
From Determinant with Rows Transposed, if you swap over two rows of a matrix, the sign of its determinant changes.
If you swap over two identical rows of a matrix, then the sign of its determinant changes from $D$, say, to $-D$.
But the matrix is the same.
So $D = -D$ and so $D = 0$.
{{qed}}
Category:Matrix Algebra
Category:Determinants
124580
124564
2013-01-08T11:53:16Z
Anghel
1607
124580
wikitext
text/x-wiki
\end{proof}
|
21692
|
\section{Square Modulo 3}
Tags: Modulo Arithmetic, Square Modulo 3, Square Numbers
\begin{theorem}
Let $x \in \Z$ be an integer.
Then one of the following holds:
{{begin-eqn}}
{{eqn | l = x^2
| o = \equiv
| r = 0 \pmod 3
| c =
}}
{{eqn | l = x^2
| o = \equiv
| r = 1 \pmod 3
| c =
}}
{{end-eqn}}
{{expand|explain exactly which}}
\end{theorem}
\begin{proof}
Let $x$ be an integer.
Using Congruence of Powers throughout, we make use of:
: $x \equiv y \pmod 3 \implies x^2 \equiv y^2 \pmod 3$
There are three cases to consider:
: $(1): \quad x \equiv 0 \pmod 3$: we have $x^2 \equiv 0^2 \pmod 3 \equiv 0 \pmod 3$
: $(2): \quad x \equiv 1 \pmod 3$: we have $x^2 \equiv 1^2 \pmod 3 \equiv 1 \pmod 3$
: $(3): \quad x \equiv 2 \pmod 3$: we have $x^2 \equiv 2^2 \pmod 3 \equiv 1 \pmod 3$
{{qed}}
\end{proof}
|
21693
|
\section{Square Modulo 3/Corollary 2}
Tags: Modulo Arithmetic, Square Modulo 3, Square Numbers
\begin{theorem}
Let $x, y, z \in \Z$ be integers.
Then:
:$x^2 + y^2 = 3 z^2 \iff x = y = z = 0$
\end{theorem}
\begin{proof}
Proof by the Method of Infinite Descent:
Suppose $u, v, w$ are the smallest non-zero positive integers such that $u^2 + v^2 = 3 w^2$.
Then from Square Modulo 3: Corollary 1 each of $u, v, w$ are multiples of $3$.
So we have $u', v', w'$ such that:
:$3 u' = u, 3 v' = v, 3 w' = w$
Then:
:$\paren {3 u'}^2 + \paren {3 v'}^2 = 3 \paren {3 w'}^2$
which leads to:
:$\paren {u'}^2 + \paren {v'}^2 = 3 \paren {w'}^2$
contradicting the supposition that $u, v, w$ are the smallest non-zero positive integers such that $u^2 + v^2 = 3 w^2$.
Thus:
: $x^2 + y^2 = 3 z^2 \implies x = y = z = 0$
If $x = y = z = 0$ then it trivially follows that $x^2 + y^2 = 3 z^2$.
{{qed}}
\end{proof}
|
21694
|
\section{Square Modulo 5}
Tags: Modulo Arithmetic, Polygonal Numbers, Square Numbers
\begin{theorem}
Let $x \in \Z$ be an integer.
Then one of the following holds:
{{begin-eqn}}
{{eqn | l = x^2
| o = \equiv
| r = 0 \pmod 5
| c =
}}
{{eqn | l = x^2
| o = \equiv
| r = 1 \pmod 5
| c =
}}
{{eqn | l = x^2
| o = \equiv
| r = 4 \pmod 5
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let $x$ be an integer.
Using Congruence of Powers throughout, we make use of $x \equiv y \pmod 5 \implies x^2 \equiv y^2 \pmod 5$.
There are five cases to consider:
: $x \equiv 0 \pmod 5$: we have $x^2 \equiv 0^2 \pmod 5 \equiv 0 \pmod 5$.
: $x \equiv 1 \pmod 5$: we have $x^2 \equiv 1^2 \pmod 5 \equiv 1 \pmod 5$.
: $x \equiv 2 \pmod 5$: we have $x^2 \equiv 2^2 \pmod 5 \equiv 4 \pmod 5$.
: $x \equiv 3 \pmod 5$: we have $x^2 \equiv 3^2 \pmod 5 \equiv 4 \pmod 5$.
: $x \equiv 4 \pmod 5$: we have $x^2 \equiv 4^2 \pmod 5 \equiv 1 \pmod 5$.
{{qed}}
\end{proof}
|
21695
|
\section{Square Modulo 5/Corollary}
Tags: Modulo Arithmetic, Square Numbers
\begin{theorem}
When written in conventional base $10$ notation, no square number ends in one of $2, 3, 7, 8$.
\end{theorem}
\begin{proof}
The absence of $2$ and $3$ from the digit that can end a square follows directly from Square Modulo 5.
As $7 \equiv 2 \pmod 5$ and $8 \equiv 3 \pmod 5$, the result for $7$ and $8$ follows directly.
{{qed}}
\end{proof}
|
21696
|
\section{Square Modulo n Congruent to Square of Inverse Modulo n}
Tags: Modulo Arithmetic
\begin{theorem}
Let $n \in \Z_{>0}$ be a (strictly) positive integer.
Then:
:$a^2 \equiv \paren {n - a}^2 \pmod n$
where the notation denotes congruence modulo $n$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {n - a}^2
| r = n^2 - 2 n - a^2
| c =
}}
{{eqn | o = \equiv
| r = a^2
| rr= \pmod n
| c =
}}
{{end-eqn}}
{{qed}}
Category:Modulo Arithmetic
\end{proof}
|
21697
|
\section{Square Number Less than One}
Tags: Real Analysis, Analysis
\begin{theorem}
Let $x$ be a real number such that $x^2 < 1$.
Then:
: $x \in \left({-1 \,.\,.\, 1}\right)$
where $\left({-1 \,.\,.\, 1}\right)$ is the open interval $\left\{{x \in \R: -1 < x < 1}\right\}$.
\end{theorem}
\begin{proof}
First note that from Square of Real Number is Non-Negative:
:$x^2 \ge 0$
From Ordering of Squares in Reals:
:$(1): \quad x > 1 \implies x^2 > 1$
:$(2): \quad x < 1 \implies x^2 < 1$
From Identity Element of Multiplication on Numbers:
:$1^2 = 1$
so it is clear that the strict inequalities apply above.
For clarity, therefore, the case where $x = \pm 1$ can be ignored.
Suppose $x \notin \left({-1 \,.\,.\, 1}\right)$.
Then either:
:$x < -1$
or:
:$x > 1$
If $x > 1$ then:
:$x^2 > 1$
If $x < -1$ then:
:$-x > 1$
From Square of Real Number is Non-Negative:
:$\left({-x}\right)^2 = x^2$
So again $x^2 > 1$.
Thus:
:$x \notin \left({-1 \,.\,.\, 1}\right) \implies x^2 > 1$
So by the Rule of Transposition:
:$x^2 < 1 \implies x \in \left({-1 \,.\,.\, 1}\right)$
{{qed}}
Category:Real Analysis
\end{proof}
|
21698
|
\section{Square Numbers are Similar Plane Numbers}
Tags: Euclidean Number Theory
\begin{theorem}
Let $a$ and $b$ be square numbers.
Then $a$ and $b$ are similar plane numbers.
\end{theorem}
\begin{proof}
Let $a = c^2$ and $b = d^2$.
We have that:
: $c : d = c : d$
The result follows by definition of similar plane numbers
{{qed}}
Category:Euclidean Number Theory
\end{proof}
|
21699
|
\section{Square Numbers which are Sum of Consecutive Powers}
Tags: 121, 400, Square Numbers
\begin{theorem}
The only two square numbers which are the sum of consecutive powers of a positive integer are $121$ and $400$:
:$121 = 3^0 + 3^1 + 3^2 + 3^3 + 3^4 = 11^2$
:$400 = 7^0 + 7^1 + 7^2 + 7^3 = 20^2$
\end{theorem}
\begin{proof}
:$121 = 1 + 3 + 9 + 27 + 81$
:$400 = 1 + 7 + 49 + 343$
{{ProofWanted|It remains to be shown that these are the only such square numbers.}}
\end{proof}
|
21700
|
\section{Square Numbers which are Sum of Sequence of Odd Cubes}
Tags: Cube Numbers, Sums of Sequences, Square Numbers
\begin{theorem}
The sequence of square numbers which can be expressed as the sum of a sequence of odd cubes from $1$ begins:
:$1, 1225, 1 \, 413 \, 721, 1 \, 631 \, 432 \, 881, \dotsc$
{{OEIS|A046177}}
The sequence of square roots of this sequence is:
:$1, 35, 1189, 40 \, 391, \dotsc$
{{OEIS|A046176}}
\end{theorem}
\begin{proof}
We have that:
{{begin-eqn}}
{{eqn | l = 1225
| r = 35^2
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^5 \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + 7^3 + 9^3
| c =
}}
{{eqn | l = 1 \, 413 \, 721
| r = 1189^2
| c =
}}
{{eqn | r = \sum_{k \mathop = 1}^{29} \paren {2 k - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + 55^3 + 57^3
| c =
}}
{{end-eqn}}
From Sum of Sequence of Odd Cubes we have:
:$\ds \sum_{j \mathop = 1}^n \paren {2 j - 1}^3 = 1^3 + 3^3 + 5^3 + \dotsb + \paren {2 n − 1}^3 = n^2 \paren {2 n^2 − 1}$
Thus we need to find all $n$ such that $2 n^2 − 1$ is square.
This corresponds to the Pell's Equation $x^2 - 2 y^2 = -1$, which has the positive integral solutions:
:$\begin {array} {r|r} x & y \\ \hline
1 & 1 \\
7 & 5 \\
41 & 29 \\
239 & 169 \\
1393 & 985 \\
\end {array}$
and so on.
By substituting $y = n$ and $x = \sqrt {2 n^2 - 1}$, we see that the products of $x$ and $y$ values give the required sequence of square roots.
{{qed}}
\end{proof}
|
21701
|
\section{Square Order 2 Matrices over Real Numbers form Ring with Unity}
Tags: Rings with Unity
\begin{theorem}
Let $S$ denote the set of square matrices of order $2$ whose entries are the set of real numbers.
Then $S$ forms a ring with unity whose unity is the matrix $\begin {pmatrix} 1 & 0 \\ 0 & 1 \end {pmatrix}$.
\end{theorem}
\begin{proof}
This is an instance of Ring of Square Matrices over Field is Ring with Unity.
{{qed}}
\end{proof}
|
21702
|
\section{Square Product of Three Consecutive Triangular Numbers}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n \times T_{n + 1} \times T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 3$
\end{theorem}
\begin{proof}
Let $T_n \times T_{n + 1} \times T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
{{begin-eqn}}
{{eqn | l = T_n \times T_{n + 1} \times T_{n + 2}
| r = \dfrac {n \paren {n + 1} } 2 \dfrac {\paren {n + 1} \paren {n + 2} } 2 \dfrac {\paren {n + 2} \paren {n + 3} } 2
| c =
}}
{{eqn | r = \dfrac {n \paren {n + 1}^2 \paren {n + 2}^2 \paren {n + 3} } 8
| c =
}}
{{eqn | r = \paren {\dfrac {\paren {n + 1} \paren {n + 2} } 2}^2 \dfrac {n \paren {n + 3} } 2
| c =
}}
{{end-eqn}}
Thus we need to find $n$ such that $\dfrac {n \paren {n + 3} } 2$ is a square number.
We see that:
{{begin-eqn}}
{{eqn | l = 3 \times \paren {3 + 3}
| r = 3 \times 6
| c =
}}
{{eqn | r = 2 \times 3^2
| c =
}}
{{end-eqn}}
Thus $n = 3$ appears to satisfy the conditions.
It remains for us to check:
{{begin-eqn}}
{{eqn | l = T_3 \times T_4 \times T_5
| r = 6 \times 10 \times 15
| c =
}}
{{eqn | r = \paren {2 \times 3} \times {2 \times 5} \times {3 \times 5}
| c =
}}
{{eqn | r = \paren {2 \times 3 \times 5}^2
| c =
}}
{{eqn | r = 30^2
| c =
}}
{{eqn | r = 900
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21703
|
\section{Square Root is Strictly Increasing}
Tags: Real Numbers, Square Roots
\begin{theorem}
The positive square root function is strictly increasing, that is:
:$ \forall x,y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$
\end{theorem}
\begin{proof}
Let $x$ and $y$ be positive real numbers such that $x < y$.
{{AimForCont}} $\sqrt x \ge \sqrt y$.
{{begin-eqn}}
{{eqn | n = 1
| l = \sqrt x
| o = \ge
| r = \sqrt y
| c =
}}
{{eqn | n = 2
| l = \sqrt x
| o = \ge
| r = \sqrt y
| c =
}}
{{eqn | l = x
| o = \ge
| r = y
| c = Real Number Axioms: $\R O2$: compatibility with multiplication, $(1) \times (2)$
}}
{{end-eqn}}
Thus a contradiction is created.
Therefore:
:$\forall x, y \in \R_{>0}: x < y \implies \sqrt x < \sqrt y$
{{qed}}
Category:Real Numbers
Category:Square Roots
\end{proof}
|
21704
|
\section{Square Root is not Lipschitz Continuous}
Tags: Counterexamples of Lipschitz Continuous Functions
\begin{theorem}
Let $\sqrt {\size x} : \R \to \R_{\ge 0}$ be a real function.
$\sqrt {\size x}$ is not Lipschitz continuous.
\end{theorem}
\begin{proof}
{{AimForCont}} $\sqrt {\size x}$ is Lipschitz continuous.
Then:
:$\exists L \in \R_{> 0} : \forall x,y \in \R : \size {\sqrt {\size x} - \sqrt {\size y} } \le L \size {x - y}$
Suppose $\ds x = \frac 1 {n^2}$ with $n \in \N$ and $y = 0$.
Then:
:$\ds \frac 1 n \le \frac L {n^2}$
In other words:
:$\forall n \in \N : n \le L$
However, $L$ was assumed to be finite.
Hence the contradiction.
{{qed}}
\end{proof}
|
21705
|
\section{Square Root is of Exponential Order Epsilon}
Tags: Exponential Order
\begin{theorem}
The positive square root function:
:$t \mapsto \sqrt t$
is of exponential order $\epsilon$ for any $\epsilon > 0$ arbitrarily small in magnitude.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \sqrt t
| o = <
| r = K e^{a t}
| c = an Ansatz
}}
{{eqn | ll = \impliedby
| l = t
| o = <
| r = \paren {K e^{a t} }^2
| c = Square Root is Strictly Increasing
}}
{{eqn | r = K^2 e^{2 a t}
| c = Exponential of Product
}}
{{eqn | r = K' e^{a' t}
| c = $K^2 = K', 2 a = a'$
}}
{{end-eqn}}
Recall from Identity is of Exponential Order Epsilon, $t < K'e^{a' t}$ for ''any'' $a' > 0$, arbitrarily small in magnitude.
Therefore the inequality $\sqrt t < K e^{a t}$ has solutions of the same nature.
{{qed}}
Category:Exponential Order
\end{proof}
|
21706
|
\section{Square Root of 2 as Sum of Egyptian Fractions}
Tags: Square Root of 2
\begin{theorem}
The square root of $2$ can be approximated by the following sequence of Egyptian fractions:
:$\sqrt 2 = 1 + \dfrac 1 3 + \dfrac 1 {13} + \dfrac 1 {253} + \dfrac 1 {218 \, 201} + \dfrac 1 {61 \, 323 \, 543 \, 802} + \cdots$
{{OEIS|A006487}}
\end{theorem}
\begin{proof}
We have by definition of the square root of $2$ that:
:$\sqrt 2 - 1 \approx 0 \cdotp 41421 \, 35623 \, 73095 \, 04880 \ldots$
By inspection:
:$\dfrac 1 3 < \sqrt 2 - 1 < \dfrac 1 2$
Thus:
:$\sqrt 2 - 1 - \dfrac 1 3 \approx 0 \cdotp 08088 \, 02290 \, 39761 \, 71546 \ldots$
Then:
:$\dfrac 1 {13} = 0 \cdotp 07692 \, 3 \ldots$ repeating
and:
:$\dfrac 1 {12} = 0 \cdotp 08333 \ldots$ repeating
and so:
:$\sqrt 2 - 1 - \dfrac 1 3 - \dfrac 1 {13} \approx 0 \cdotp 00395 \, 71521 \, 16684 \, 79239 \cdots$
Thus one can generate this sequence of denominators $\sequence {D_n}$ by:
:$\ds D_n = \ceiling {\paren {\sqrt 2 - \sum_{i \mathop = 0}^{n - 1} \frac 1 {D_i}}^{-1}}$
\end{proof}
|
21707
|
\section{Square Root of 2 is Algebraic of Degree 2}
Tags: Square Root of 2, Algebraic Numbers
\begin{theorem}
The square root of $2$ is an algebraic number of degree $2$.
\end{theorem}
\begin{proof}
Suppose $\sqrt 2$ could be expressed as the root of the linear polynomial:
:$a_1 x + a_0 = 0$
for some $a_0, a_1 \in \Q$.
Then:
:$\sqrt 2 = -\dfrac {a_0} {a_1}$
and would be rational.
But as Square Root of 2 is Irrational, this is not the case.
However, $\sqrt 2$ is a root of the polynomial of degree $2$:
:$x^2 - 2 = 0$
Hence the result by definition of degree of algebraic number.
{{qed}}
\end{proof}
|
21708
|
\section{Square Root of 2 is Irrational}
Tags: Square Root of 2 is Irrational, Examples of Square Roots, Irrationality Proofs, Square Root of 2, Number Theory, Proofs by Contradiction
\begin{theorem}
:$\sqrt 2$ is irrational.
\end{theorem}
\begin{proof}
be definition <math>2 | p^2 \iff \exists q \in \mathbb{Z}</math> such that <math>2q = p^2 </math>
This theorem is then proved by Theorem A
\end{proof}
|
21709
|
\section{Square Root of Complex Number in Cartesian Form}
Tags: Complex Analysis, Square Roots, Complex Numbers
\begin{theorem}
Let $z \in \C$ be a complex number.
Let $z = x + i y$ where $x, y \in \R$ are real numbers.
Let $z$ not be wholly real, that is, such that $y \ne 0$.
Then the square root of $z$ is given by:
:$z^{1/2} = \pm \paren {a + i b}$
where:
{{begin-eqn}}
{{eqn | l = a
| r = \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}
| c =
}}
{{eqn | l = b
| r = \frac y {\cmod y} \sqrt {\frac {-x + \sqrt {x^2 + y^2} } 2}
| c =
}}
{{end-eqn}}
\end{theorem}
\begin{proof}
Let $a + i b \in z^{1/2}$.
Then:
{{begin-eqn}}
{{eqn | l = \paren {a + i b}^2
| r = x + i y
| c = {{Defof|Square Root|subdef = Complex Number|index = 4|Square Root of Complex Number}}
}}
{{eqn | n = 1
| ll= \leadsto
| l = a^2 + 2 i a b - b^2
| r = x + i y
| c = Square of Sum and $i^2 = -1$
}}
{{end-eqn}}
Equating imaginary parts in $(1)$:
{{begin-eqn}}
{{eqn | l = 2 a b
| r = y
| c =
}}
{{eqn | n = 2
| ll= \leadsto
| l = b
| r = \frac y {2 a}
| c = rearranging
}}
{{end-eqn}}
Equating real parts in $(1)$:
{{begin-eqn}}
{{eqn | l = a^2 - b^2
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = a^2 - \paren {\frac y {2 a} }^2
| r = x
| c = substituting for $b$ from $(2)$
}}
{{eqn | ll= \leadsto
| l = 4 a^4 - 4 a^2 x - y^2
| r = 0
| c = multiplying by $4 a^2$ and rearranging
}}
{{eqn | ll= \leadsto
| l = a^2
| r = \frac {4 x \pm \sqrt {16 x^2 + 16 y^2} } 8
| c = Quadratic Formula
}}
{{eqn | ll= \leadsto
| l = a^2
| r = \frac {x \pm \sqrt {x^2 + y^2} } 2
| c = dividing top and bottom by $4$
}}
{{eqn | n = 3
| ll= \leadsto
| l = a
| r = \pm \sqrt {\frac {x + \sqrt {x^2 + y^2} } 2}
| c = taking the square root
}}
{{end-eqn}}
Note that in $(3)$, only the positive square root of the discriminant $x^2 + y^2$ is used.
This is because the negative square root of $x^2 + y^2$ would yield $\dfrac {x - \sqrt {x^2 + y^2} } 2 < 0$.
As $a \in \R$, it is necessary that $\dfrac {x + \sqrt {x^2 + y^2} } 2 > 0$.
Hence $\sqrt {x^2 + y^2} > 0$.
Then:
{{begin-eqn}}
{{eqn | l = b
| r = \frac y {2 a}
| c = from $(2)$
}}
{{eqn | r = \frac y {2 \paren {\pm \sqrt {\dfrac {x + \sqrt {x^2 + y^2} } 2} } }
| c =
}}
{{eqn | ll= \leadsto
| l = b^2
| r = \frac {y^2} {2 \paren {x + \sqrt {x^2 + y^2} } }
| c =
}}
{{eqn | ll= \leadsto
| l = b^2
| r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x + \sqrt {x^2 + y^2} } \paren {x - \sqrt {x^2 + y^2} } }
| c = multiplying top and bottom by $x - \sqrt {x^2 + y^2}$
}}
{{eqn | ll= \leadsto
| l = b^2
| r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {2 \paren {x^2 - \paren {x^2 + y^2} } }
| c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
| l = b^2
| r = \frac {y^2 \paren {x - \sqrt {x^2 + y^2} } } {- 2 y^2}
| c =
}}
{{eqn | ll= \leadsto
| l = b^2
| r = \frac {-x + \sqrt {x^2 + y^2} } 2
| c =
}}
{{end-eqn}}
But from $(2)$ we have:
:$b = \dfrac y {2 a}$
and so having picked either the positive square root or negative square root of either $a^2$ or $b^2$, the root of the other is forced.
So:
:if $y > 0$, then $a$ and $b$ are both of the same sign.
Thus:
:$b = 1 \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
:if $y < 0$, then $a$ and $b$ are of opposite sign.
Thus:
:$b = \paren {-1} \times \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
Hence:
:$b = \dfrac y {\cmod y} \sqrt {\dfrac {-x + \sqrt {x^2 + y^2} } 2}$
{{qed}}
\end{proof}
|
21710
|
\section{Square Root of Number Minus Square Root/Proof 1}
Tags: Square Root of Number Plus or Minus Square Root
\begin{theorem}
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
{{:Square Root of Number Plus Square Root}}
\end{theorem}
\begin{proof}
We are given that $a^2 - b > 0$.
Then:
:$a > \sqrt b$
and so $\ds \sqrt {a - \sqrt b}$ is defined on the real numbers.
Let $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Observe that:
:$\ds 0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$
Squaring both sides gives:
{{begin-eqn}}
{{eqn | l = a - \sqrt b
| r = \paren {\sqrt x - \sqrt y}^2
| c =
}}
{{eqn | r = x + y - 2 \sqrt {x y}
| c =
}}
{{end-eqn}}
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
From $\sqrt b = 2 \sqrt {x y}$ we get:
{{begin-eqn}}
{{eqn | l = \sqrt b
| r = 2 \sqrt {x y}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = b
| r = 4 x y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x y
| r = \frac b 4
| c =
}}
{{end-eqn}}
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
:$z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
Because we have that $x > y$:
{{begin-eqn}}
{{eqn | l = x = z_1
| r = \dfrac {a + \sqrt {a^2 - b} } 2
}}
{{eqn | l = y = z_2
| r = \dfrac {a - \sqrt {a^2 - b} } 2
}}
{{end-eqn}}
Subsituting into $\ds \sqrt {a - \sqrt b} = \sqrt x - \sqrt y$:
{{begin-eqn}}
{{eqn | l = \sqrt {a - \sqrt b}
| r = \sqrt x - \sqrt y
| c =
}}
{{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| c =
}}
{{end-eqn}}
{{qed}}
Category:Square Root of Number Plus or Minus Square Root
\end{proof}
|
21711
|
\section{Square Root of Number Minus Square Root/Proof 2}
Tags: Square Root of Number Plus or Minus Square Root
\begin{theorem}
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
{{:Square Root of Number Minus Square Root}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2
| r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 - 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| c = multiplying out
}}
{{eqn | r = a - \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }
| c = simplifying
}}
{{eqn | r = a - \sqrt {a^2 - \paren {a^2 - b} }
| c = Difference of Two Squares
}}
{{eqn | r = a - \sqrt b
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} - \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| r = \sqrt {a - \sqrt b}
| c = taking square root of both sides
}}
{{end-eqn}}
{{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}}
{{qed}}
Category:Square Root of Number Plus or Minus Square Root
\end{proof}
|
21712
|
\section{Square Root of Number Plus Square Root/Proof 1}
Tags: Square Root of Number Plus or Minus Square Root
\begin{theorem}
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
{{:Square Root of Number Plus Square Root}}
\end{theorem}
\begin{proof}
We are given that $a^2 - b > 0$.
Then:
:$a > \sqrt b$
and so $\ds \sqrt {a + \sqrt b}$ is defined on the real numbers.
Let $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Squaring both sides gives:
{{begin-eqn}}
{{eqn | l = a + \sqrt b
| r = \paren {\sqrt x + \sqrt y}^2
| c =
}}
{{eqn | r = x + y + 2 \sqrt {x y}
| c =
}}
{{end-eqn}}
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
{{explain|How do you know that the $a$ and $b$ which are $x + y$ and $2 \sqrt {x y}$ are the same $a$ and $b$ that you started with?
$x$ and $y$ are free to choose. They were introduced by hand, and can be set to any value provided that they satisfy the constraint above. This is similar to the proof of Cardano's Formula.
In that case it needs to be explained. As it stands, it looks as though $x$ and $y$ are pulled out of thin air, with no actual indication that having picked them, they bear the relations given to $a$ and $b$ as presented. It's incredibly confusing for beginners, and others whose abilities and understanding are limited, like me.}}
From $\sqrt b = 2 \sqrt {x y}$ we get:
{{begin-eqn}}
{{eqn | l = \sqrt b
| r = 2 \sqrt {x y}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = b
| r = 4 x y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x y
| r = \frac b 4
| c =
}}
{{end-eqn}}
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
:$z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
{{explain|the notation $z_{1, 2}$}}
{{WLOG}}:
{{begin-eqn}}
{{eqn | l = x = z_1
| r = \dfrac {a + \sqrt {a^2 - b} } 2
}}
{{eqn | l = y = z_2
| r = \dfrac {a - \sqrt {a^2 - b} } 2
}}
{{end-eqn}}
Subsituting into $\ds \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:
{{begin-eqn}}
{{eqn | l = \sqrt {a + \sqrt b}
| r = \sqrt x + \sqrt y
| c =
}}
{{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| c =
}}
{{end-eqn}}
{{qed}}
Category:Square Root of Number Plus or Minus Square Root
\end{proof}
|
21713
|
\section{Square Root of Number Plus Square Root/Proof 2}
Tags: Square Root of Number Plus or Minus Square Root
\begin{theorem}
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
{{:Square Root of Number Plus Square Root}}
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {\sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2} }^2
| r = \dfrac {a + \sqrt {a^2 - b} } 2 + \dfrac {a - \sqrt {a^2 - b} } 2 + 2 \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| c = multiplying out
}}
{{eqn | r = a + \sqrt {a + \sqrt {a^2 - b} } \sqrt {a - \sqrt {a^2 - b} }
| c = simplifying
}}
{{eqn | r = a + \sqrt {a^2 - \paren {a^2 - b} }
| c = Difference of Two Squares
}}
{{eqn | r = a + \sqrt b
| c = simplifying
}}
{{eqn | ll= \leadsto
| l = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| r = \sqrt {a + \sqrt b}
| c = taking square root of both sides
}}
{{end-eqn}}
{{finish|Report on the matter of the signs and magnitudes of $a$ and $b$ according to the constraints given}}
{{qed}}
Category:Square Root of Number Plus or Minus Square Root
\end{proof}
|
21714
|
\section{Square Root of Number Plus or Minus Square Root}
Tags: Square Root of Number Plus or Minus Square Root, Square Roots
\begin{theorem}
Let $a$ and $b$ be (strictly) positive real numbers such that $a^2 - b > 0$.
Then:
\end{theorem}
\begin{proof}
We are given that $a^2 - b > 0$.
Then:
:$a > \sqrt b$
and so both $\displaystyle \sqrt {a + \sqrt b}$ and $\displaystyle \sqrt {a - \sqrt b}$ are defined on the real numbers.
Let $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$ where $x, y$ are (strictly) positive real numbers.
Squaring both sides gives:
{{begin-eqn}}
{{eqn | l = a + \sqrt b
| r = \paren {\sqrt x + \sqrt y}^2
| c =
}}
{{eqn | r = x + y + 2 \sqrt {x y}
| c =
}}
{{end-eqn}}
Set $x + y = a$ and $\sqrt b = 2 \sqrt {x y}$
{{explain|How do you know that the $a$ and $b$ which are $x + y$ and $2 \sqrt {x y}$ are the same $a$ and $b$ that you started with?
$x$ and $y$ are free to choose. They were introduced by hand, and can be set to any value provided that they satisfy the constraint above. This is similar to the proof of Cardano's Formula.
In that case it needs to be explained. As it stands, it looks as though $x$ and $y$ are pulled out of thin air, with no actual indication that having picked them, they bear the relations given to $a$ and $b$ as presented. It's incredibly confusing for beginners, and others whose abilities and understanding are limited, like me.}}
From $\sqrt b = 2 \sqrt {x y}$ we get:
{{begin-eqn}}
{{eqn | l = \sqrt b
| r = 2 \sqrt {x y}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = b
| r = 4 x y
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = x y
| r = \frac b 4
| c =
}}
{{end-eqn}}
By Viète's Formulas, $x$ and $y$ are solutions to the quadratic equation:
:$z^2 - a z + \dfrac b 4 = 0$
From Solution to Quadratic Equation:
:$z_{1, 2} = \dfrac {a \pm \sqrt {a^2 - b} } 2$
where $a^2 - b > 0$ (which is a given)
{{explain|the notation $z_{1, 2}$}}
{{WLOG}}:
{{begin-eqn}}
{{eqn | l = x = z_1
| r = \dfrac {a + \sqrt {a^2 - b} } 2
}}
{{eqn | l = y = z_2
| r = \dfrac {a - \sqrt {a^2 - b} } 2
}}
{{end-eqn}}
Subsituting into $\displaystyle \sqrt {a + \sqrt b} = \sqrt x + \sqrt y$:
{{begin-eqn}}
{{eqn | l = \sqrt {a + \sqrt b}
| r = \sqrt x + \sqrt y
| c =
}}
{{eqn | r = \sqrt {\dfrac {a + \sqrt {a^2 - b} } 2} + \sqrt {\dfrac {a - \sqrt {a^2 - b} } 2}
| c =
}}
{{end-eqn}}
which completes the proof of $(1)$.
For the proof of $(2)$, the same style of proof is followed, observing that:
:$0 < \sqrt {a - \sqrt b} = \sqrt x - \sqrt y \implies x > y$
{{finish}}
\end{proof}
|
21715
|
\section{Square Root of Prime is Irrational}
Tags: Square Root of Prime is Irrational, Square Roots, Irrationality Proofs, Number Theory, Proofs by Contradiction, Prime Numbers
\begin{theorem}
The square root of any prime number is irrational.
\end{theorem}
\begin{proof}
Let $p$ be prime.
Suppose that $\sqrt p$ is rational.
Then there exist natural numbers $m$ and $n$ such that:
{{begin-eqn}}
{{eqn | l = \sqrt p
| r = \frac m n
| c =
}}
{{eqn | ll= \leadsto
| l = p
| r = \frac {m^2} {n^2}
| c =
}}
{{eqn | ll= \leadsto
| l = n^2 p
| r = m^2
| c =
}}
{{end-eqn}}
Any prime in the prime factorizations of $n^2$ and $m^2$ must occur an even number of times because they are squares.
Thus, $p$ must occur in the prime factorization of $n^2 p$ an odd number of times.
Therefore, $p$ occurs as a factor of $m^2$ an odd number of times, a contradiction.
So $\sqrt p$ must be irrational.
{{qed}}
\end{proof}
|
21716
|
\section{Square Root of Sum as Sum of Square Roots}
Tags: Square Root of Sum as Sum of Square Roots, Algebra
\begin{theorem}
Let $a, b \in \R, a \ge b$.
Then:
:$\sqrt {a + b} = \sqrt {\dfrac a 2 + \dfrac {\sqrt {a^2 - b^2}} 2} + \sqrt {\dfrac a 2 - \dfrac {\sqrt {a^2 - b^2}} 2}$
\end{theorem}
\begin{proof}
Let $\sqrt {a + b}$ be expressed in the form $\sqrt c + \sqrt d$.
From Square of Sum:
: $a + b = c + d + 2 \sqrt {c d}$
We now need to solve the simultaneous equations:
: $a = c + d$
: $b = 2 \sqrt {c d}$
First:
{{begin-eqn}}
{{eqn | l = a
| r = c + d
}}
{{eqn | n = 1
| ll= \implies
| l = d
| r = a - c
| c = subtracting $c$ from both sides
}}
{{end-eqn}}
Solving for $c$:
{{begin-eqn}}
{{eqn | l = b
| r = 2 \sqrt {c d}
}}
{{eqn | ll= \implies
| l = b^2
| r = 4 c d
| c = squaring both sides
}}
{{eqn | r = 4 c \left({a - c}\right)
| c = substituting $d = a - c$ from $(1)$
}}
{{eqn | r = 4 a c - 4 c^2
| c = Real Multiplication Distributes over Addition
}}
{{eqn | ll= \implies
| l = 4 c^2 - 4 a c + b^2
| r = 0
| c = adding $4 c^2 - 4 a c$ to both sides
}}
{{eqn | n = 2
| ll= \implies
| l = c
| r = \frac a 2 \pm \frac {\sqrt {a^2 - b^2} } 2
| c = Quadratic Formula
}}
{{end-eqn}}
Solving for $d$:
{{begin-eqn}}
{{eqn | l = d
| r = a - c
}}
{{eqn | r = a - \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2
| c = substituting $c = \dfrac a 2 \pm \dfrac {\sqrt {a^2 - b^2} } 2$ from $(2)$
}}
{{eqn | r = \frac a 2 \mp \frac {\sqrt {a^2 - b^2} } 2
}}
{{end-eqn}}
From Real Addition is Commutative, the sign of the square root may be chosen arbitrarily, provided opposite signs are chosen for $c$ and $d$.
{{qed}}
Category:Algebra
194775
193845
2014-09-26T07:58:22Z
Ybab321
1675
194775
wikitext
text/x-wiki
\end{proof}
|
21717
|
\section{Square Sum of Three Consecutive Triangular Numbers}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
Let $T_n$ denote the $n$th triangular number for $n \in \Z_{>0}$ a (strictly) positive integer.
Let $T_n + T_{n + 1} + T_{n + 2}$ be a square number.
Then at least one value of $n$ fulfils this condition:
:$n = 5$
\end{theorem}
\begin{proof}
Let $T_n + T_{n + 1} + T_{n + 2} = m^2$ for some $m \in \Z_{>0}$.
We have:
{{begin-eqn}}
{{eqn | l = T_n + T_{n + 1} + T_{n + 2}
| r = \dfrac {n \paren {n + 1} } 2 + \dfrac {\paren {n + 1} \paren {n + 2} } 2 + \dfrac {\paren {n + 2} \paren {n + 3} } 2
| c =
}}
{{eqn | r = \dfrac {n \paren {n + 1} + \paren {n + 1} \paren {n + 2} + \paren {n + 2} \paren {n + 3} } 8
| c =
}}
{{eqn | r = \dfrac {n^2 + n + n^2 + 3 n + 2 + n^2 + 5 n + 6} 2
| c = multiplying out
}}
{{eqn | r = \dfrac {3 n^2 + 9 n + 8} 2
| c = multiplying out
}}
{{end-eqn}}
Thus we need to find $n$ such that $\dfrac {3 n^2 + 9 n + 8} 2$ is a square number.
We see that:
{{begin-eqn}}
{{eqn | l = \dfrac {3 \times 5^2 + 9 \times 5 + 8} 2
| r = \dfrac {75 + 45 + 8} 2
| c =
}}
{{eqn | r = \dfrac {128} 2
| c =
}}
{{eqn | r = 64
| c =
}}
{{end-eqn}}
Thus $n$ appears to satisfy the conditions.
It remains for us to check:
{{begin-eqn}}
{{eqn | l = T_5 + T_6 + T_7
| r = 15 + 21 + 28
| c =
}}
{{eqn | r = 64
| c =
}}
{{eqn | r = 8^2
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21718
|
\section{Square of 1 Less than Number Base}
Tags: Square of 1 Less than Number Base, Square Numbers
\begin{theorem}
Let $b \in \Z$ be an integer such that $b > 2$.
Let $n = b - 1$.
The square of $n$ is expressed in base $b$ as:
:$n^2 = \left[{c1}\right]_b$
where $c = b - 2$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = n^2
| r = \left({b - 1}\right)^2
| c =
}}
{{eqn | r = b^2 - 2 b + 1
| c =
}}
{{eqn | r = b \left({b - 2}\right) + 1
| c =
}}
{{end-eqn}}
The result follows by definition of number base.
{{qed}}
\end{proof}
|
21719
|
\section{Square of Complex Conjugate is Complex Conjugate of Square}
Tags: Complex Conjugates, Complex Multiplication
\begin{theorem}
Let $z \in \C$ be a complex number.
Let $\overline z$ denote the complex conjugate of $z$.
Then:
: $\overline {z^2} = \left({\overline z}\right)^2$
\end{theorem}
\begin{proof}
A direct consequence of Product of Complex Conjugates:
: $\overline {z_1 z_2} = \overline {z_1} \cdot \overline {z_2}$
for two complex numbers $z_1, z_2 \in \C$.
{{qed}}
\end{proof}
|
21720
|
\section{Square of Complex Modulus equals Complex Modulus of Square}
Tags: Complex Modulus, Complex Multiplication
\begin{theorem}
Let $z \in \C$ be a complex number.
Let $\cmod z$ be the modulus of $z$.
Then:
: $\cmod {z^2} = \cmod z^2$
\end{theorem}
\begin{proof}
From Complex Modulus of Product of Complex Numbers:
: $\cmod {z_1 z_2} = \cmod {z_1} \cmod {z_2}$
for $z_1, z_2 \in \C$.
Set $z = z_1 = z_2$ and the result follows.
{{qed}}
\end{proof}
|
21721
|
\section{Square of Coprime Number is Coprime}
Tags: Coprime Integers, Coprime
\begin{theorem}
Let $a$ and $b$ be coprime integers:
:$a, b \in \Z: a \perp b$
Then:
:$a^2 \perp b$
{{:Euclid:Proposition/VII/25}}
\end{theorem}
\begin{proof}
Let $a \perp b$.
Let $a^2 = c$.
Let $d = a$.
As $a \perp b$ it follows that $d \perp b$.
From {{EuclidPropLink|book=VII|prop=24|title=Integer Coprime to all Factors is Coprime to Whole}}:
: $a d \perp b$
But $a d = c = a^2$.
Hence the result.
{{qed}}
{{Euclid Note|25|VII}}
\end{proof}
|
21722
|
\section{Square of Covariance is Less Than or Equal to Product of Variances}
Tags: Variance
\begin{theorem}
Let $X$ and $Y$ be random variables.
Let the variances of $X$ and $Y$ exist and be finite.
Then:
:$\paren {\cov {X, Y} }^2 \le \var X \, \var Y$
where $\cov {X, Y}$ denotes the covariance of $X$ and $Y$.
\end{theorem}
\begin{proof}
We have, by the definition of variance, that both:
:$\expect {\paren {X - \expect X}^2}$
and:
:$\expect {\paren {Y - \expect Y}^2}$
exist and are finite.
Therefore:
{{begin-eqn}}
{{eqn | l = \paren {\cov {X, Y} }^2
| r = \paren {\expect {\paren {X - \expect X} \paren {Y - \expect Y} } }^2
| c = {{Defof|Covariance}}
}}
{{eqn | o = \le
| r = \expect {\paren {X - \expect X}^2} \expect {\paren {Y - \expect Y}^2}
| c = Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares
}}
{{eqn | r = \var X \, \var Y
| c = {{Defof|Variance}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21723
|
\section{Square of Cube Number is Cube}
Tags: Euclidean Number Theory, Square of Cube Number is Cube
\begin{theorem}
Let $a \in \N$ be a natural number.
Let $a$ be a cube number.
Then $a^2$ is also a cube number.
{{:Euclid:Proposition/IX/3}}
\end{theorem}
\begin{proof}
By the definition of cube number:
:$\exists k \in \N: k^3 = a$
Thus:
{{begin-eqn}}
{{eqn | l = a^2
| r = \left({k^3}\right)^2
| c =
}}
{{eqn | r = k^6
| c =
}}
{{eqn | r = \left({k^2}\right)^3
| c =
}}
{{end-eqn}}
Thus:
:$\exists r = k^2 \in \N: a = r^3$
Hence the result by definition of cube number.
{{qed}}
{{Euclid Note|3|IX|The proof as given here is not that given by Euclid, as the latter is unwieldy and of limited use.}}
\end{proof}
|
21724
|
\section{Square of Element Less than Unity in Ordered Integral Domain}
Tags: Integral Domains, Ordered Integral Domains
\begin{theorem}
Let $\struct {D, +, \times, \le}$ be an ordered integral domain.
Let $x \in D$ such that $0 < x < 1$.
Then:
: $0 < x \times x < x$
\end{theorem}
\begin{proof}
We have that $0 < x < 1$.
From Relation Induced by Strict Positivity Property is Compatible with Multiplication:
:$0 \times x < x \times x < 1 \times x$
Hence the result.
{{qed}}
Category:Ordered Integral Domains
\end{proof}
|
21725
|
\section{Square of Elements of Standard Ordered Basis equals 1}
Tags: Standard Ordered Bases, Dot Product, Standard Ordered Basis
\begin{theorem}
Let $\tuple {\mathbf i, \mathbf j, \mathbf k}$ be the standard ordered basis of Cartesian $3$-space $S$.
Then:
:$\mathbf i^2 = \mathbf j^2 = \mathbf k^2 = 1$
where $\mathbf i^2$ and so on denotes the square of a vector quantity:
:$\mathbf i^2 := \mathbf i \cdot \mathbf i$
\end{theorem}
\begin{proof}
By definition, the Cartesian $3$-space is a frame of reference consisting of a rectangular coordinate system.
By definition of Component of Vector in $3$-space, the vectors $\mathbf i$, $\mathbf j$ and $\mathbf k$ are the unit vectors in the direction of the $x$-axis, $y$-axis and $z$-axis respectively.
Hence $\mathbf i^2$ is the square of a unit vector:
:$\mathbf i^2 = \norm {\mathbf i}^2 = 1^2 = 1$
and the same for $\mathbf j^2$ and $\mathbf k^2$.
{{qed}}
\end{proof}
|
21726
|
\section{Square of Expectation of Product is Less Than or Equal to Product of Expectation of Squares}
Tags: Expectation
\begin{theorem}
Let $X$ and $Y$ be random variables.
Let the expectation of $X Y$, $\expect {X Y}$, exist and be finite.
Then:
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
\end{theorem}
\begin{proof}
Note that:
:$\map \Pr {Y^2 \ge 0} = 1$
so we have by Expectation of Non-Negative Random Variable is Non-Negative:
:$\expect {Y^2} \ge 0$
First, take $\expect {Y^2} > 0$.
Let $Z$ be a random variable with:
:$Z = X - Y \dfrac {\expect {X Y} } {\expect {Y^2} }$
Note that we have:
:$\map \Pr {Z^2 \ge 0} = 1$
so again applying Expectation of Non-Negative Random Variable is Non-Negative, we have:
:$\expect {Z^2} \ge 0$
That is:
{{begin-eqn}}
{{eqn | l = 0
| o = \le
| r = \expect {Z^2}
}}
{{eqn | r = \expect {\paren {X - Y \frac {\expect {X Y} } {\expect {Y^2} } }^2}
}}
{{eqn | r = \expect {X^2 - 2 X Y \frac {\expect {X Y} } {\expect {Y^2} } + Y^2 \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2} }
| c = Square of Sum
}}
{{eqn | r = \expect {X^2} - 2 \expect {X Y} \frac {\expect {X Y} } {\expect {Y^2} } + \expect {Y^2} \frac {\paren {\expect {X Y} }^2} {\paren {\expect {Y^2} }^2}
| c = Linearity of Expectation Function
}}
{{eqn | r = \expect {X^2} - \frac {\paren {\expect {X Y} }^2 } {\expect {Y^2} }
}}
{{end-eqn}}
We therefore have:
:$\dfrac {\paren {\expect {X Y} }^2} {\expect {Y^2} } \le \expect {X^2}$
giving:
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
as required.
It remains to address the case $\expect {Y^2} = 0$.
Note that since $\map \Pr {Y^2 \ge 0} = 1$, from Condition for Expectation of Non-Negative Random Variable to be Zero we necessarily have:
:$\map \Pr {Y^2 = 0} = 1$
That is:
:$\map \Pr {Y = 0} = 1$
This implies that the random variable $X Y$ has:
:$\map \Pr {X Y = 0} = 1$
From which, we have that:
:$\map \Pr {X Y \ge 0} = 1$
So, applying Expectation of Non-Negative Random Variable is Non-Negative again we have:
:$\expect {X Y} = 0$
With that, we have:
:$\paren {\expect {X Y} }^2 = 0$
and:
:$\expect {X^2} \expect {Y^2} = \expect {X^2} \times 0 = 0$
So the inequality:
:$\paren {\expect {X Y} }^2 \le \expect {X^2} \expect {Y^2}$
also holds in the case $\expect {Y^2} = 0$, completing the proof.
{{qed}}
\end{proof}
|
21727
|
\section{Square of Metric does not necessarily form Metric}
Tags: Examples of Metric Spaces
\begin{theorem}
Let $M = \struct {A, d}$ be a metric space.
Let $d_4: A^2 \to \R$ be the mapping defined as:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \paren {\map d {x, y} }^2$
Then $d_4$ may or may not be a metric for $A$.
\end{theorem}
\begin{proof}
Let $d$ be the standard discrete metric on $M$.
Then:
:$\forall \tuple {x, y} \in A^2: \map {d_4} {x, y} = \map d {x, y}$
and indeed in this case $d_4$ is a metric for $A$.
{{qed|lemma}}
Let $M = \struct {\R, d}$ be the real number line with the usual (Euclidean) metric.
Let $x = 1$, $y = \dfrac 1 2$ and $z = 0$.
We have:
{{begin-eqn}}
{{eqn | l = \map {d_4} {x, y}
| r = \size {1 - \dfrac 1 2}^2
| c =
}}
{{eqn | r = \dfrac 1 4
| c =
}}
{{eqn | l = \map {d_4} {y, z}
| r = \size {\dfrac 1 2 - 0}^2
| c =
}}
{{eqn | r = \dfrac 1 4
| c =
}}
{{eqn | l = \map {d_4} {x, z}
| r = \size {1 - 0}^2
| c =
}}
{{eqn | r = 1
| c =
}}
{{end-eqn}}
So we have:
:$\map {d_4} {x, y} + \map {d_4} {y, z} < \map {d_4} {x, z}$
and it is seen that $d_4$ does not satisfy {{Metric-space-axiom|2}}.
{{qed}}
\end{proof}
|
21728
|
\section{Square of Non-Zero Element of Ordered Integral Domain is Strictly Positive}
Tags: Integral Domains, Ordered Integral Domains, Orderings
\begin{theorem}
Let $\struct {D, +, \times, \le}$ be an ordered integral domain whose zero is $0_D$.
Then:
:$\forall x \in D: x \ne 0_D \iff \map P {x \times x}$
where $\map P {x \times x}$ denotes that $x \times x$ has the (strict) positivity property.
That is, the square of any element of an ordered integral domain is (strictly) positive {{iff}} that element is non-zero.
\end{theorem}
\begin{proof}
Let $x = 0_D$.
Then $x \times x = 0_D \times 0_D = 0_D$ by the properties of the ring zero.
Thus by definition of strict positivity property:
:$\neg \map P {0_D \times 0_D}$
{{qed|lemma}}
Now suppose $x \ne 0_D$.
One of two cases applies:
:$\map P x$
:$\neg \map P x$
Let $\map P x$.
Then by definition of (strict) positivity:
:$\map P {x \times x}$
Now suppose $\neg \map P x$.
Then by the trichotomy law of ordered integral domains:
:$\map P {-x}$
Then again by definition:
:$\map P {\paren {-x} \times \paren {-x} }$.
But by Product of Ring Negatives:
:$\paren {-x} \times \paren {-x} = x \times x$
So again:
:$\map P {x \times x}$
Hence the result.
{{qed}}
\end{proof}
|
21729
|
\section{Square of Non-Zero Real Number is Strictly Positive}
Tags: Real Numbers
\begin{theorem}
:$\forall x \in \R: x \ne 0 \implies x^2 > 0$
\end{theorem}
\begin{proof}
There are two cases to consider:
:$(1): \quad x > 0$
:$(2): \quad x < 0$
Let $x > 0$.
Then:
{{begin-eqn}}
{{eqn | l = x \times x
| o = >
| r = 0
| c = Strictly Positive Real Numbers are Closed under Multiplication
}}
{{end-eqn}}
Let $x < 0$.
Then:
{{begin-eqn}}
{{eqn | l = x
| o = <
| r = 0
| c =
}}
{{eqn | ll= \leadsto
| l = x \times x
| o = >
| r = x \times 0
| c = Real Number Ordering is Compatible with Multiplication: Negative Factor
}}
{{eqn | r = 0
| c = Real Zero is Zero Element
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21730
|
\section{Square of Number Always Exists}
Tags: Numbers
\begin{theorem}
Let $x$ be a number.
Then its square $x^2$ is guaranteed to exist.
\end{theorem}
\begin{proof}
Whatever flavour of number under discussion, the algebraic structure $\struct {\mathbb K, +, \times}$ in which this number sits is at least a semiring.
The binary operation that is multiplication is therefore closed on that algebraic structure.
Therefore:
: $\forall x \in \mathbb K: x \times x \in \mathbb K$
{{qed}}
Category:Numbers
\end{proof}
|
21731
|
\section{Square of Odd Number as Difference between Triangular Numbers}
Tags: Triangular Numbers, Square Numbers
\begin{theorem}
Let $n \in \Z_{\ge 0}$ be a positive integer.
Then:
:$\exists a, b \in \Z_{\ge 0}: \paren {2 n + 1}^2 = T_a - T_b$
where:
:$T_a$ and $T_b$ are triangular numbers
:$T_a$ and $T_b$ are coprime.
That is, the square of every odd number is the difference between two coprime triangular numbers.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = T_a - T_b
| r = \dfrac {a^2 + a} 2 - \dfrac {b^2 + b} 2
| c = Closed Form for Triangular Numbers
}}
{{end-eqn}}
Let $a = 3b + 1$
{{begin-eqn}}
{{eqn | l = T_{3 b + 1} - T_b
| r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1} 2 - \dfrac {b^2 + b} 2
| c =
}}
{{eqn | r = \dfrac {\paren {3 b + 1}^2 + 3 b + 1 - b^2 + b} 2
| c =
}}
{{eqn | r = \dfrac {9 b^2 + 6 b + 1 + 3 b + 1 - b^2 - b} 2
| c =
}}
{{eqn | r = \dfrac {8 b^2 + 8 b + 2} 2
| c =
}}
{{eqn | r = 4 b^2 + 4 b + 1
| c =
}}
{{eqn | r = \paren {2 b + 1}^2
| c =
}}
{{end-eqn}}
The square of every odd number can be made through this method.
{{begin-eqn}}
{{eqn | l = T_{3 b + 1}
| r = \dfrac {\paren {3 b + 1} ^ 2 + 3 b + 1} 2
| c =
}}
{{eqn | r = \dfrac {9 b^2 + 9 b + 2} 2
| c =
}}
{{eqn | r = 9 \paren {\dfrac {b^2 + b} 2} + 1
| c =
}}
{{eqn | ll= \leadsto
| l = T_{3 b + 1} - 9 T_b
| r = 1
| c =
}}
{{eqn | ll= \leadsto
| l = \gcd \set {T_{3 b + 1}, T_b}
| r = 1
| c = Bézout's Lemma
}}
{{end-eqn}}
Both triangular numbers are coprime.
{{qed}}
\end{proof}
|
21732
|
\section{Square of Ones Matrix}
Tags: Matrix Algebra
\begin{theorem}
Let $\mathbf J = \sqbrk 1_n$ be a square ones matrix of order $n$.
Then $\mathbf J^2 = n \mathbf J$.
That is:
:$\begin{bmatrix}
1 & 1 & \cdots & 1 \\
1 & 1 & \cdots & 1 \\
\vdots & \vdots & \ddots & \vdots \\
1 & 1 & \cdots & 1
\end{bmatrix}^2 = \begin{bmatrix}
n & n & \cdots & n \\
n & n & \cdots & n \\
\vdots & \vdots & \ddots & \vdots \\
n & n & \cdots & n
\end{bmatrix}$
\end{theorem}
\begin{proof}
Follows directly from the definition of matrix multiplication:
:$\ds \forall i \in \closedint 1 m, j \in \closedint 1 p: c_{i j} = \sum_{k \mathop = 1}^n a_{i k} \circ b_{k j}$
In this case, $m = n$ and $a_{i k} = b_{k j} = 1$.
Hence:
:$\ds c_{i j} = \sum_{k \mathop = 1}^n 1 \times 1 = n$
{{qed}}
Category:Matrix Algebra
\end{proof}
|
21733
|
\section{Square of Pythagorean Prime is Hypotenuse of Pythagorean Triangle}
Tags: Pythagorean Triangles
\begin{theorem}
Let $p$ be a Pythagorean prime.
Then $p^2$ is the hypotenuse of a Pythagorean triangle.
\end{theorem}
\begin{proof}
By Fermat's Two Squares Theorem, a Pythagorean prime, $p$ can be expressed in the form:
:$p = m^2 + n^2$
where $m$ and $n$ are (strictly) positive integers.
:$(1): \quad m \ne n$, otherwise $p$ would be of the form $2 m^2$ and so even and therefore not a prime.
:$(2): \quad m \perp n$, otherwise $\exists c \in \Z: p = c^2 \paren {p^2 + q^2}$ for some $p, q \in \Z$ and therefore not a prime.
:$(3): \quad m$ and $n$ are of opposite parity, otherwise $p = m^2 + n^2$ is even and therefore not a prime.
{{WLOG}}, let $m > n$.
From Solutions of Pythagorean Equation, the triple $\tuple {2 m n, m^2 - n^2, m^2 + n^2}$ forms a primitive Pythagorean triple such that $m^2 + n^2$ forms the hypotenuse of a primitive Pythagorean triangle.
Thus we have that $p^2 = r^2 + s^2$ for some (strictly) positive integers $r$ and $s$.
Similarly, we have that $r \ne s$, otherwise $p^2$ would be of the form $2 r^2$ and so not a square number.
{{WLOG}}, let $r > s$.
From Solutions of Pythagorean Equation, the triple $\tuple {2 r s, r^2 - s^2, r^2 + s^2}$ forms a Pythagorean triple such that $p^2 = r^2 + s^2$ forms the hypotenuse of a Pythagorean triangle.
{{qed}}
\end{proof}
|
21734
|
\section{Square of Quadratic Gauss Sum}
Tags: Number Theory
\begin{theorem}
Let $p$ be an odd prime.
Let $a$ be an integer coprime to $p$.
Let $\map g {a, p}$ denote the quadratic Gauss sum of $a$ and $p$.
Then:
:$\map g {a, p}^2 = \paren {\dfrac {-1} p} \cdot p$
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Number Theory
\end{proof}
|
21735
|
\section{Square of Random Variable with t-Distribution has F-Distribution}
Tags: Student's t-Distribution, F-Distribution
\begin{theorem}
Let $k$ be a strictly positive integer.
Let $X \sim t_k$ where $t_k$ is the $t$-distribution with $k$ degrees of freedom.
Then:
:$X^2 \sim F_{1, k}$
where $F_{1, k}$ is the $F$-distribution with $\tuple {1, k}$ degrees of freedom.
\end{theorem}
\begin{proof}
Let $Y \sim F_{1, k}$.
We aim to show that:
:$\map \Pr {Y < x^2} = \map \Pr {|X| < x}$
for all $x \ge 0$.
That is:
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all $x \ge 0$.
We have:
{{begin-eqn}}
{{eqn | l = \map \Pr {Y < x^2}
| r = \int_0^{x^2} \frac {k^{k/2} 1^{1/2} u^{\paren {1/2} - 1} } {\paren {k + u}^{\paren {1 + k}/2} \map \Beta {1/2, k/2} } \rd u
| c = {{Defof|F-Distribution}}
}}
{{eqn | r = \frac {k^{k/2} } {k^{k/2} \sqrt k \map \Beta {1/2, k/2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
| c = extracting a factor of $k^{-\paren {1 + k}/2}$, rewriting
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt k \map \Gamma {\frac 1 2} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
| c = {{Defof|Beta Function}}
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^{x^2} \frac 1 {\sqrt u} \paren {1 + \frac u k}^{-\paren {1 + k}/2} \rd u
| c = Gamma Function of One Half
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x \frac {2 v} {\sqrt {v^2} } \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
| c = substituting $u = v^2$
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_0^x 2 \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
}}
{{eqn | r = \frac {\map \Gamma {\frac {k + 1} 2} } {\sqrt {\pi k} \map \Gamma {\frac k 2} } \int_{-x}^x \paren {1 + \frac {v^2} k}^{-\paren {1 + k}/2} \rd v
| c = Definite Integral of Even Function
}}
{{eqn | r = \map \Pr {-x < X < x}
| c = {{Defof|Student's t-Distribution}}
}}
{{end-eqn}}
{{qed}}
Category:Student's t-Distribution
Category:F-Distribution
\end{proof}
|
21736
|
\section{Square of Real Number is Non-Negative}
Tags: Real Analysis
\begin{theorem}
Let $x \in \R$.
Then:
: $0 \le x^2$
\end{theorem}
\begin{proof}
There are two cases to consider:
: $(1): \quad x = 0$
: $(2): \quad x \ne 0$
Let $x = 0$.
Then:
:$x^2 = 0$
and thus
:$0 \le x^2$
{{qed|lemma}}
Let $x \ne 0$.
From Square of Non-Zero Real Number is Strictly Positive it follows that:
:$0 < x^2$
and so by definition:
:$0 \le x^2$
{{Qed}}
\end{proof}
|
21737
|
\section{Square of Repunit times Sum of Digits}
Tags: Repunits, Square Numbers
\begin{theorem}
The following pattern emerges:
{{begin-eqn}}
{{eqn | l = 121 \times \paren {1 + 2 + 1}
| r = 22^2
}}
{{eqn | l = 12 \, 321 \times \paren {1 + 2 + 3 + 2 + 1}
| r = 333^2
}}
{{eqn | l = 1 \, 234 \, 321 \times \paren {1 + 2 + 3 + 4 + 3 + 2 + 1}
| r = 4444^2
}}
{{end-eqn}}
and so on, up until $999 \, 999 \, 999^2$ after which the pattern breaks down.
\end{theorem}
\begin{proof}
From Square of Repunit:
{{begin-eqn}}
{{eqn | l = 121
| r = 11^2
}}
{{eqn | l = 12 \, 321
| r = 111^2
}}
{{eqn | l = 1 \, 234 \, 321
| r = 1111^2
}}
{{end-eqn}}
and so on.
Then from 1+2+...+n+(n-1)+...+1 = n^2:
{{begin-eqn}}
{{eqn | l = 1 + 2 + 1
| r = 2^2
}}
{{eqn | l = 1 + 2 + 3 + 2 + 1
| r = 3^2
}}
{{eqn | l = 1 + 2 + 3 + 4 + 3 + 2 + 1
| r = 4^2
}}
{{end-eqn}}
and so on.
Then:
{{begin-eqn}}
{{eqn | l = 11^2 \times 2^2
| r = 22^2
}}
{{eqn | l = 111^2 \times 3^2
| r = 333^2
}}
{{eqn | l = 1111^2 \times 4^2
| r = 4444^2
}}
{{end-eqn}}
The pattern breaks down after $9$:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
{{qed}}
\end{proof}
|
21738
|
\section{Square of Riemann Zeta Function}
Tags: Riemann Zeta Function, Zeta Function
\begin{theorem}
:$\ds \map {\zeta^2} z = \sum_{k \mathop = 1}^\infty \frac {\map {\sigma_0} k} {k^z}$
where:
:$\zeta$ is the Riemann zeta function
:$\sigma_0$ is the divisor counting function.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \map {\zeta^2} z
| r = \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^z} } \paren {\sum_{n \mathop = 1}^\infty \frac 1 {n^z} }
| c =
}}
{{eqn | r = \paren {1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots} \paren {1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots}
| c =
}}
{{end-eqn}}
Expanding this product, we get:
{{begin-eqn}}
{{eqn | r = 1 + \frac 1 {2^z} + \frac 1 {3^z} + \frac 1 {4^z} + \frac 1 {5^z} + \frac 1 {6^z} + \cdots
| c =
}}
{{eqn | ro= +
| o =
| r = \frac 1 {2^z} + \frac 1 {4^z} + \frac 1 {6^z} + \frac 1 {8^z} + \frac 1 {10^z} + \frac 1 {12^z} + \cdots
| c =
}}
{{eqn | ro= +
| o =
| r = \frac 1 {3^z} + \frac 1 {6^z} + \frac 1 {9^z} + \frac 1 {12^z} + \frac 1 {15^z} + \frac 1 {18^z} + \cdots
| c =
}}
{{eqn | ro= +
| o =
| r = \frac 1 {4^z} + \frac 1 {8^z} + \frac 1 {12^z} + \frac 1 {16^z} + \frac 1 {20^z} + \frac 1 {24^z} + \cdots
| c =
}}
{{eqn | ro= +
| o =
| r = \frac 1 {5^z} + \frac 1 {10^z} + \frac 1 {15^z} + \frac 1 {20^z} + \frac 1 {25^z} + \frac 1 {30^z} + \cdots
| c =
}}
{{eqn | ro= +
| o =
| r = \frac 1 {6^z} + \frac 1 {12^z} + \frac 1 {18^z} + \frac 1 {24^z} + \frac 1 {30^z} + \frac 1 {36^z} + \cdots
| c =
}}
{{eqn | ro= \vdots
| o =
| c =
}}
{{eqn | r = 1 + \frac 2 {2^z} + \frac 2 {3^z} + \frac 3 {4^z} + \frac 2 {5^z} + \frac 4 {6^z} + \cdots
| c =
}}
{{end-eqn}}
We see that each $\dfrac 1 {n^z}$ term in this sum will occur as many times as there are ways represent $n$ as $a b$, counting order.
But this is precisely the number of divisors of $n$, since each way of representing $n = a b$ corresponds to the first term $a$ of the product.
Hence this sum is:
:$\ds \sum_{n \mathop = 1}^\infty \frac {\map {\sigma_0} n} {z^n}$
as desired.
{{qed}}
Category:Riemann Zeta Function
\end{proof}
|
21739
|
\section{Square of Small-Digit Palindromic Number is Palindromic}
Tags: Recreational Mathematics, Square of Small-Digit Palindromic Number is Palindromic, Palindromic Numbers, Square Numbers
\begin{theorem}
Let $n$ be an integer such that the sum of the squares of the digits of $n$ in decimal representation is less than $10$.
Let $n$ be palindromic.
Then $n^2$ is also palindromic.
The sequence of such numbers begins:
:$0, 1, 2, 3, 11, 22, 101, 111, 121, 202, 212, 1001, 1111, \dots$
{{OEIS|A057135}}
\end{theorem}
\begin{proof}
Let $\ds n = \sum_{k \mathop = 0}^m a_k 10^k$ be a number satisfying the conditions above.
Then:
{{begin-eqn}}
{{eqn | n = 1
| l = \sum_{k \mathop = 0}^m a_k^2
| o = <
| r = 10
}}
{{eqn | n = 2
| l = a_k
| r = a_{m - k}
| rr = \forall k: 0 \le k \le m
}}
{{end-eqn}}
Consider $\ds n^2 = \paren {\sum_{k \mathop = 0}^m a_k 10^k}^2$.
From definition of Multiplication of Polynomials, the coefficient of $10^l$ in the product is:
{{begin-eqn}}
{{eqn | o =
| r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k
}}
{{eqn | o = \le
| r = \sqrt {\paren {\sum_{j \mathop = 0}^l a_j^2} \paren {\sum_{k \mathop = 0}^l a_k^2} }
| c = Cauchy's Inequality
}}
{{eqn | r = \sum_{j \mathop = 0}^l a_j^2
}}
{{eqn | o = \le
| r = \sum_{j \mathop = 0}^m a_j^2
}}
{{eqn | o = <
| r = 10
| c = From (1)
}}
{{end-eqn}}
so no carries occur in the multiplication, and this form satisfies Basis Representation Theorem.
Moreover:
{{begin-eqn}}
{{eqn | l = \sum_{\substack {j \mathop + k \mathop = 2 m \mathop - l \\ j, k \mathop \in \Z} } a_j a_k
| r = \sum_{\substack {m \mathop - j \mathop + m \mathop - k \mathop = 2 m \mathop - l \\ m \mathop - j, m \mathop - k \mathop \in \Z} } a_{m - j} a_{m - k}
}}
{{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_{m - j} a_{m - k}
}}
{{eqn | r = \sum_{\substack {j \mathop + k \mathop = l \\ j, k \mathop \in \Z} } a_j a_k
| c = From (2)
}}
{{end-eqn}}
so the coefficient of $10^{2 m - l}$ is equal to the coefficient of $10^l$ in the expansion of $n^2$.
This shows that $n^2$ is palindromic.
{{qed}}
\end{proof}
|
21740
|
\section{Square of Small Repunit is Palindromic}
Tags: Repunits, Square Numbers
\begin{theorem}
The squares of repunits with up to $9$ digits are palindromic.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 1^2
| r = 1
}}
{{eqn | l = 11^2
| r = 121
| c =
}}
{{eqn | l = 111^2
| r = 12 \, 321
| c =
}}
{{eqn | l = 1111^2
| r = 1 \, 234 \, 321
| c =
}}
{{eqn | l = 11 \, 111^2
| r = 123 \, 454 \, 321
| c =
}}
{{eqn | l = 111 \, 111^2
| r = 12 \, 345 \, 654 \, 321
| c =
}}
{{eqn | l = 1 \, 111 \, 111^2
| r = 1 \, 234 \, 567 \, 654 \, 321
| c =
}}
{{eqn | l = 11 \, 111 \, 111^2
| r = 123 \, 456 \, 787 \, 654 \, 321
| c =
}}
{{eqn | l = 111 \, 111 \, 111^2
| r = 12 \, 345 \, 678 \, 987 \, 654 \, 321
| c =
}}
{{end-eqn}}
but:
:$1 \, 111 \, 111 \, 111^2 = 1 \, 234 \, 567 \, 900 \, 987 \, 654 \, 321$
{{qed}}
\end{proof}
|
21741
|
\section{Square of Standard Gaussian Random Variable has Chi-Squared Distribution}
Tags: Chi-Squared Distribution, Gaussian Distribution
\begin{theorem}
Let $X \sim \Gaussian 0 1$ where $\Gaussian 0 1$ is the standard Gaussian distribution.
Then $X^2 \sim \chi^2_1$ where $\chi^2_1$ is the chi-square distribution with $1$ degree of freedom.
\end{theorem}
\begin{proof}
Let $Y \sim \chi^2_1$.
We aim to show that:
:$\map \Pr {Y < x^2} = \map \Pr {-x < X < x}$
for all real $x \ge 0$.
We have:
{{begin-eqn}}
{{eqn | l = \map \Pr {Y < x^2}
| r = \frac 1 {\sqrt 2 \map \Gamma {1 / 2} } \int_0^{x^2} t^{\paren {1 / 2} - 1} e^{-t / 2} \rd t
| c = {{Defof|Chi-Squared Distribution}}
}}
{{eqn | r = \frac 1 {\sqrt {2 \pi} } \int_0^{x^2} \frac {e^{-t / 2} } {\sqrt t} \rd t
| c = Gamma Function of One Half
}}
{{eqn | r = \frac 2 {\sqrt {2 \pi} } \int_0^x u \frac {e^{-u^2 / 2} } {\sqrt {u^2} } \rd u
| c = substituting $t = u^2$
}}
{{eqn | r = \frac 2 {\sqrt {2 \pi} } \int_0^x e^{-u^2 / 2} \rd u
}}
{{eqn | r = \frac 1 {\sqrt {2 \pi} } \int_{-x}^x e^{-u^2 / 2} \rd u
| c = Definite Integral of Even Function
}}
{{eqn | r = \map \Pr {-x < X < x}
| c = {{Defof|Gaussian Distribution}}
}}
{{end-eqn}}
{{qed}}
Category:Gaussian Distribution
Category:Chi-Squared Distribution
\end{proof}
|
21742
|
\section{Square of Tangent Minus Square of Sine}
Tags: Trigonometric Identities
\begin{theorem}
:$\tan^2 x - \sin^2 x = \tan^2 x \ \sin^2 x$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l=\tan^2 x - \sin^2 x
| r=\frac {\sin^2 x} {\cos^2x} - \sin^2 x
| c=Tangent is Sine divided by Cosine
}}
{{eqn | r=\frac {\sin^2 x - \sin^2 x \ \cos^2 x} {\cos^2x}
}}
{{eqn | r=\frac{\sin^2 x \left({1 - \cos^2 x}\right)} {\cos^2 x}
}}
{{eqn | r=\tan^2 x \left({1 - \cos^2 x}\right)
| c=Tangent is Sine divided by Cosine
}}
{{eqn | r=\tan^2 x \ \sin^2 x
| c=Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
21743
|
\section{Square of Vandermonde Matrix}
Tags: Vandermonde Matrices, Vandermonde Matrix
\begin{theorem}
The square of the Vandermonde matrix of order $n$:
: $\mathbf V = \begin{bmatrix}
x_1 & x_2 & \cdots & x_n \\
x_1^2 & x_2^2 & \cdots & x_n^2 \\
\vdots & \vdots & \ddots & \vdots \\
x_1^n & x_2^n & \cdots & x_n^n
\end{bmatrix}$
is symmetrical in $x_1, \ldots, x_n$.
{{questionable|The case $n {{=}} 2$ left me clueless to what could possibly be intended here; only $\mathbf {V V}^T$ is trivially seen symmetric in the $x_n$, but this can hardly be called a square}}
\end{theorem}
\begin{proof}
{{proof wanted}}
Category:Vandermonde Matrices
\end{proof}
|
21744
|
\section{Square of Vector Quantity in Coordinate Form}
Tags: Dot Product
\begin{theorem}
Let $\mathbf a$ be a vector in a vector space $\mathbf V$ of $n$ dimensions:
$\ds \mathbf a = \sum_{k \mathop = 1}^n a_k \mathbf e_k$
where $\tuple {\mathbf e_1, \mathbf e_2, \ldots, \mathbf e_n}$ is the standard ordered basis of $\mathbf V$.
Then:
:$\ds \mathbf a^2 = \sum_{k \mathop = 1}^n a_k^2$
where $\mathbf a^2$ denotes the square of $\mathbf a$.
\end{theorem}
\begin{proof}
By definition of square of $\mathbf a$:
:$\mathbf a^2 = \mathbf a \cdot \mathbf a$
By definition of dot product:
:$\ds \mathbf a \cdot \mathbf a = a_1 a_1 + a_2 a_2 + \cdots + a_n a_n = \sum_{k \mathop = 1}^n a_k^2$
{{qed}}
\end{proof}
|
21745
|
\section{Square of n such that 2n-1 is Composite is not Sum of Square and Prime}
Tags: Prime Numbers, Numbers not Sum of Square and Prime, Square Numbers
\begin{theorem}
Let $n^2$ be a square such that $2 n - 1$ is composite.
Then $n^2$ cannot be expressed as the sum of a square and a prime.
\end{theorem}
\begin{proof}
The case where $n = 1$ is trivial, as there are no prime numbers less than $1$.
Let $n, m \in \Z$ be integers such that $n > 1$.
Let $n^2 = m^2 + p$ where $p$ is prime.
Then:
{{begin-eqn}}
{{eqn | l = p
| r = n^2 - m^2
| c =
}}
{{eqn | r = \paren {n + m} \paren {n - m}
| c = Difference of Two Squares
}}
{{eqn | ll= \leadsto
| l = n - m
| r = 1
| c = {{Defof|Prime Number}}
}}
{{eqn | ll= \leadsto
| l = m
| r = n - 1
| c = {{Defof|Prime Number}}
}}
{{eqn | ll= \leadsto
| l = n + m
| r = 2 n - 1
| c =
}}
{{eqn | r = p
| c =
}}
{{end-eqn}}
So if $2 n - 1$ is composite, there exists no prime $p$ such that $n^2 = m^2 + p$.
Further, if such a $p$ does exist, then $m = n - 1$, and so:
:$n^2 = \paren {n - 1}^2 + p$
{{qed}}
Category:Numbers not Sum of Square and Prime
\end{proof}
|
21746
|
\section{Square on Medial Straight Line/Lemma}
Tags: Euclidean Number Theory
\begin{theorem}
{{:Euclid:Proposition/X/22/Lemma}}
Algebraically:
:$a : b = a^2 : a b$
\end{theorem}
\begin{proof}
:400px
Let $FE$ and $EG$ be straight lines.
Let the square $DF$ be described on $FE$.
Let the rectangle $GD$ be completed.
From Areas of Triangles and Parallelograms Proportional to Base:
:$FE : EG = FD : DG$
We have that $DG$ is the rectangle contained by $DE$ and $EG$.
But as $DF$ is a square, then $DE = FE$.
Thus $DG$ is the rectangle contained by $FE$ and $EG$.
So as $FE$ is to $EG$, so is the square on $EF$ to the rectangle contained by $FE$ and $EG$.
{{qed}}
{{Euclid Note|22|X}}
\end{proof}
|
21747
|
\section{Square which is Difference between Square and Square of Reversal}
Tags: Reversals, Square Numbers
\begin{theorem}
$33^2 = 65^2 - 56^2$
This is the only square of a $2$-digit number which has this property.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = 33^2
| r = 1089
| c =
}}
{{eqn | l = 65^2 - 56^2
| r = 4225 - 3136
| c =
}}
{{eqn | r = 1089
| c =
}}
{{end-eqn}}
Let $\sqbrk {xy}$ be a $2$-digit number such that $x \le y$ and $\sqbrk {xy}^2 - \sqbrk {yx}^2$ is a square of a $2$-digit number.
The case $x = y$ gives the solution $\sqbrk {xy}^2 - \sqbrk {yx}^2 = 0$, which is not a square of a $2$-digit number.
For $x \ne y$:
{{begin-eqn}}
{{eqn | l = \sqbrk {xy}^2 - \sqbrk {yx}^2
| r = \paren {10 x + y}^2 - \paren {10 y + x}^2
}}
{{eqn | r = 100 x^2 + 20 x y + y^2 - 100 y^2 - 20 y x - x^2
}}
{{eqn | r = 99 \paren {x^2 - y^2}
}}
{{eqn | r = 3^2 \times 11 \paren {x - y} \paren {x + y}
}}
{{eqn | ll = \leadsto
| l = \paren {x - y} \paren {x + y}
| r = 11 n^2
| c = for some integer $n$
}}
{{end-eqn}}
By Euclid's Lemma for Prime Divisors, one of $x - y, x + y$ must be divisible by $11$.
Hence from Absolute Value of Integer is not less than Divisors, either $x - y$ or $x + y$ must be greater than or equal to $11$.
Since $x - y < x < 9$, $x - y$ cannot be a multiple of $11$.
From $1 = 1 + 0 \le x + y < 9 + 9 = 18$, we have that $x + y = 11$.
This implies that $x - y$ is a square number.
$x + y = 11$ gives $\tuple {x,y} = \tuple {6,5}, \tuple {7,4}, \tuple {8,3}, \tuple {9,2}$ as the possible solutions.
Among these solutions, only $\tuple {6,5}$ has a difference of a square number: $1$.
Therefore the only square of a $2$-digit number expressible as a difference between a square and the square of its reversal is:
:$65^2 - 56^2 = 33^2$.
{{qed}}
\end{proof}
|
21748
|
\section{Square whose Perimeter equals its Area}
Tags: Squares
\begin{theorem}
The $4 \times 4$ square is the only square whose area in square units equals its perimeter in units.
The area and perimeter of this square are $16$.
\end{theorem}
\begin{proof}
Let $S$ be a square whose area equals its perimeter.
Let $A$ be the area of $S$.
Let $P$ be the perimeter of $S$.
Let $b$ be the length of one side of $S$.
From Area of Square:
:$A = b^2$
From Perimeter of Rectangle:
:$P = 2 b + 2 b = 4 b$
Setting $A = P$
:$b^2 = 4 b$
and so:
:$b = 4$
and so:
:$A = 16 = P$
{{qed}}
\end{proof}
|
21749
|
\section{Squares Ending in Repeated Digits}
Tags: Square Numbers
\begin{theorem}
A square number $n^2$ can end in a repeated digit {{iff}} either:
:$(1): \quad n^2$ is a multiple of $100$, in which case $n$ is a multiple of $10$
:$(2): \quad n^2$ ends in $44$ and $n$ ends in $12, 38, 62$ or $88$.
\end{theorem}
\begin{proof}
Let $n \in \Z_{>0}$ end in $a b$.
By the Basis Representation Theorem, $n$ can be expressed as:
:$n = 100 k + 10 a + b$
for some $k \in \Z_{>0}$ and for $0 \le a < 10, 0 \le b < 10$.
Then:
{{begin-eqn}}
{{eqn | l = n^2
| r = \paren {100k + 10 a + b}^2
| c =
}}
{{eqn | r = 100^2 k^2 + 2000 k a + 200 k b + 100 b^2 + 20 a b + b^2
| c =
}}
{{eqn | o = \equiv
| r = 20 a b + b^2
| rr = \pmod {100}
}}
{{end-eqn}}
Thus the nature of the last $2$ digits of $n^2$ are not dependent upon $k$.
So we can ignore all digits of $n$ except the last $2$.
Note that if $b = 0$ we have that $b^2 = 0$ and $20 a b = 0$.
So the last $2$ digits of the square of a multiple of $10$ are both $0$, and $n^2$ is a multiple of $100$.
Let $a = 5 + c$ where $0 \le c < 5$.
We have that:
{{begin-eqn}}
{{eqn | l = \paren {10 a + b}^2
| r = \paren {10 \paren {5 + c} + b}^2
| c =
}}
{{eqn | r = \paren {10 c + 50 + b}^2
| c =
}}
{{eqn | r = 100 c^2 + 2 \times 500 c + 2 \times 10 b c + 2 \times 50 b + b^2
| c =
}}
{{eqn | r = 100 c^2 + 1000 c + 100 b + 20 b c + b^2
| c =
}}
{{eqn | r = 1000 c + 100 b + \paren {10 c + b}^2
| c =
}}
{{eqn | o = \equiv
| r = \paren {10 c + b}^2
| rr = \pmod {100}
}}
{{end-eqn}}
So the square of a number ending in $c b$, where $0 \le c < 5$, ends in the same $2$ digits as the square a number ending in $a b$ where $a = c + 5$.
It remains to list the integers from $1$ to $49$, generating their squares and investigating their last $2$ digits:
{{begin-eqn}}
{{eqn | l = 01^2
| r = 01
}}
{{eqn | l = 02^2
| r = 04
}}
{{eqn | l = 03^2
| r = 09
}}
{{eqn | l = 04^2
| r = 16
}}
{{eqn | l = 05^2
| r = 25
}}
{{eqn | l = 06^2
| r = 36
}}
{{eqn | l = 07^2
| r = 49
}}
{{eqn | l = 08^2
| r = 64
}}
{{eqn | l = 09^2
| r = 81
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 11^2
| r = 121
}}
{{eqn | l = 12^2
| r = 144
| c = $n^2$ ends in $44$ and $n$ ends in $12$
}}
{{eqn | l = 13^2
| r = 169
}}
{{eqn | l = 14^2
| r = 196
}}
{{eqn | l = 15^2
| r = 225
}}
{{eqn | l = 16^2
| r = 256
}}
{{eqn | l = 17^2
| r = 289
}}
{{eqn | l = 18^2
| r = 324
}}
{{eqn | l = 19^2
| r = 361
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 21^2
| r = 441
}}
{{eqn | l = 22^2
| r = 484
}}
{{eqn | l = 23^2
| r = 529
}}
{{eqn | l = 24^2
| r = 576
}}
{{eqn | l = 25^2
| r = 225
}}
{{eqn | l = 26^2
| r = 676
}}
{{eqn | l = 27^2
| r = 729
}}
{{eqn | l = 28^2
| r = 784
}}
{{eqn | l = 29^2
| r = 841
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 31^2
| r = 961
}}
{{eqn | l = 32^2
| r = 1024
}}
{{eqn | l = 33^2
| r = 1089
}}
{{eqn | l = 34^2
| r = 1156
}}
{{eqn | l = 35^2
| r = 1225
}}
{{eqn | l = 36^2
| r = 1296
}}
{{eqn | l = 37^2
| r = 1369
}}
{{eqn | l = 38^2
| r = 1444
| c = $n^2$ ends in $44$ and $n$ ends in $38$
}}
{{eqn | l = 39^2
| r = 1521
}}
{{end-eqn}}
{{begin-eqn}}
{{eqn | l = 41^2
| r = 1681
}}
{{eqn | l = 42^2
| r = 1764
}}
{{eqn | l = 43^2
| r = 1849
}}
{{eqn | l = 44^2
| r = 1936
}}
{{eqn | l = 45^2
| r = 2025
}}
{{eqn | l = 46^2
| r = 2116
}}
{{eqn | l = 47^2
| r = 2209
}}
{{eqn | l = 48^2
| r = 2304
}}
{{eqn | l = 49^2
| r = 2401
}}
{{end-eqn}}
It is seen that the only $n^2$ ending in a repeated digit end in $44$.
The corresponding $n$ are seen to be $12$ and $38$.
Adding $5$ to the $10$s digit of each gives us $62$ and $88$ as further such:
{{begin-eqn}}
{{eqn | l = 62^2
| r = 3844
}}
{{eqn | l = 88^2
| r = 7744
}}
{{end-eqn}}
The result follows.
{{qed}}
\end{proof}
|
21750
|
\section{Squares Ending in n Occurrences of m-Digit Pattern}
Tags: Number Theory, Squares Ending in n Occurrences of m-Digit Pattern, Recreational Mathematics
\begin{theorem}
Suppose there exists some integer $x$ such that $x^2$ ends in some $m$-digit pattern ending in an odd number not equal to $5$ and is preceded by another odd number, i.e.:
:$\exists x \in \Z: x^2 \equiv \sqbrk {1 a_1 a_2 \cdots a_m} \pmod {2 \times 10^m}$
where $a_m$ is odd, $a_m \ne 5$ and $m \ge 1$.
Then for any $n \ge 1$, there exists some integer with not more than $m n$-digits such that its square ends in $n$ occurrences of the $m$-digit pattern.
\end{theorem}
\begin{proof}
We prove that there exists a sequence $\sequence {b_n}$ with the properties:
:$b_n < 10^{m n}$
:$b_n^2 \equiv \underbrace {\sqbrk {1 \paren {a_1 \cdots a_m} \cdots \paren {a_1 \cdots a_m}}}_{n \text { occurrences}} \pmod {2 \times 10^{m n}}$
by induction:
\end{proof}
|
21751
|
\section{Squares of Diagonals of Parallelogram}
Tags: Parallelograms
\begin{theorem}
Let $ABCD$ be a parallelogram.
:400px
Then:
:$AC^2 + BD^2 = AB^2 + BC^2 + CD^2 + DA^2$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = AC^2
| r = AB^2 + BC^2 - 2 (AB) (BC) \cos \angle B
| c = Cosine Rule
}}
{{eqn | l = BD^2
| r = BC^2 + CD^2 - 2 (CD) (BC) \cos \angle C
| c = Cosine Rule
}}
{{eqn | r = DA^2 + CD^2 - 2 (AB) (CD) \cos \angle C
| c = as $AB = CD$ and $BC = DA$
}}
{{eqn | ll= \leadsto
| l = AC^2 + BD^2
| r = AB^2 + BC^2 + DA^2 + CD^2 - 2 (AB) (BC) \paren {\cos \angle B + \cos \angle C}
| c =
}}
{{end-eqn}}
But we have that $\angle C$ and $\angle B$ are supplementary:
:$\angle C = \angle B = 180 \degrees$
Thus from Cosine of Supplementary Angle:
:$\cos \angle B + \cos \angle C = 0$
The result follows.
{{Qed}}
\end{proof}
|
21752
|
\section{Squares of Linear Combination of Sine and Cosine}
Tags: Trigonometric Identities
\begin{theorem}
:$\paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2 = a^2 + b^2$
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | l = \paren {a \cos x + b \sin x}^2 + \paren {b \cos x - a \sin x}^2
| r = a^2 \cos^2 x + 2 a b \cos x \ \sin x + b^2 \sin^2 x
}}
{{eqn | o =
| ro= +
| r = b^2 \cos^2 x - 2 a b \sin x \ \cos x + a^2 \sin^2 x
}}
{{eqn | r = \paren {a^2 + b^2} \paren {\sin^2 x + \cos^2 x}
}}
{{eqn | r = a^2 + b^2
| c = Sum of Squares of Sine and Cosine
}}
{{end-eqn}}
{{qed}}
Category:Trigonometric Identities
\end{proof}
|
21753
|
\section{Squares whose Digits form Consecutive Decreasing Integers}
Tags: Recreational Mathematics, Square Numbers
\begin{theorem}
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive decreasing integers begins:
:$91, 9079, 9901, 733 \, 674, 999 \, 001, 88 \, 225 \, 295, 99 \, 990 \, 001, \ldots$
{{OEIS|A030467}}
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 91^2
| r = 8281
| c =
}}
{{eqn | l = 9079^2
| r = 82 \, 428 \, 241
| c =
}}
{{eqn | l = 9901^2
| r = 98 \, 029 \, 801
| c =
}}
{{eqn | l = 733 \, 674^2
| r = 538 \, 277 \, 538 \, 276
| c =
}}
{{eqn | l = 999 \, 001^2
| r = 998 \, 002 \, 998 \, 001
| c =
}}
{{end-eqn}}
They can be determined by inspection.
{{qed}}
\end{proof}
|
21754
|
\section{Squares whose Digits form Consecutive Increasing Integers}
Tags: Recreational Mathematics, Square Numbers
\begin{theorem}
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive increasing integers begins:
:$428, 573, 727, 846, 7810, 36 \, 365, 63 \, 636, 326 \, 734, \ldots$
{{OEIS|A030467}}
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 428^2
| r = 183 \, 184
| c =
}}
{{eqn | l = 573^2
| r = 328 \, 329
| c =
}}
{{eqn | l = 727^2
| r = 528 \, 529
| c =
}}
{{eqn | l = 846^2
| r = 715 \, 716
| c =
}}
{{eqn | l = 7810^2
| r = 6099 \, 6100
| c =
}}
{{eqn | l = 36 \, 365^2
| r = 13224 \, 13225
| c =
}}
{{eqn | l = 63 \, 636^2
| r = 40495 \, 40496
| c =
}}
{{eqn | l = 326 \, 734^2
| r = 106755 \, 106756
| c =
}}
{{end-eqn}}
They can be determined by inspection.
{{qed}}
\end{proof}
|
21755
|
\section{Squares whose Digits form Consecutive Integers}
Tags: Recreational Mathematics, Square Numbers
\begin{theorem}
The sequence of integers whose squares have a decimal representation consisting of the concatenation of $2$ consecutive integers, either increasing or decreasing begins:
:$91, 428, 573, 727, 846, 7810, 9079, 9901, 36 \, 365, 63 \, 636, 326 \, 734, 673 \, 267, 733 \, 674, \ldots$
This sequence can be divided into two subsequences:
Those where the consecutive integers are increasing:
:$428, 573, 727, 846, 7810, 36 \, 365, 63 \, 636, 326 \, 734, 673 \, 267, \ldots$
{{OEIS|A030467}}
Those where the consecutive integers are decreasing:
:$91, 9079, 9901, 733 \, 674, 999 \, 001, 88 \, 225 \, 295, \ldots$
{{OEIS|A054216}}
\end{theorem}
\begin{proof}
We have:
{{begin-eqn}}
{{eqn | l = 91^2
| r = 8281
| c =
}}
{{eqn | l = 428^2
| r = 183 \, 184
| c =
}}
{{eqn | l = 573^2
| r = 328 \, 329
| c =
}}
{{eqn | l = 727^2
| r = 528 \, 529
| c =
}}
{{eqn | l = 846^2
| r = 715 \, 716
| c =
}}
{{eqn | l = 7810^2
| r = 6099 \, 6100
| c =
}}
{{eqn | l = 9079^2
| r = 82 \, 428 \, 241
| c =
}}
{{eqn | l = 9901^2
| r = 98 \, 029 \, 801
| c =
}}
{{eqn | l = 36 \, 365^2
| r = 13224 \, 13225
| c =
}}
{{eqn | l = 63 \, 636^2
| r = 40495 \, 40496
| c =
}}
{{eqn | l = 326 \, 734^2
| r = 106755 \, 106756
| c =
}}
{{end-eqn}}
They can be determined by inspection.
{{qed}}
\end{proof}
|
21756
|
\section{Squares with All Odd Digits}
Tags: Number Theory, Square Numbers
\begin{theorem}
The only squares whose digits<ref>That is, when written in base 10 notation.</ref> are all odd are $1$ and $9$.
\end{theorem}
\begin{proof}
If $n$ is even, then at least the last digit of $n^2$ is even.
So for $n^2$ to have its digits all odd, $n$ itself must be odd.
We can see immediately that $1 = 1^2$ and $9 = 3^2$ fit the criterion.
Of the other 1-digit odd integers, we have $5^2 = 25, 7^2 = 49, 9^2 = 81$, all of which have an even digit.
Now, let $n > 10$ be an odd integer. There are five cases to consider:
* $n = 10 p + 1$: we have $\left({10p + 1}\right)^2 = 100 p^2 + 20 p + 1 = 10 \left({10 p^2 + 2 p}\right) + 1$.
* $n = 10 p + 3$: we have $\left({10p + 3}\right)^2 = 100 p^2 + 60 p + 9 = 10 \left({10 p^2 + 6 p}\right) + 9$.
* $n = 10 p + 5$: we have $\left({10p + 5}\right)^2 = 100 p^2 + 100 p + 25 = 10 \left({10 p^2 + 10 p + 2}\right)+ 5$.
* $n = 10 p + 7$: we have $\left({10p + 7}\right)^2 = 100 p^2 + 140 p + 49 = 10 \left({10 p^2 + 14 p + 4}\right)+ 9$.
* $n = 10 p + 9$: we have $\left({10p + 9}\right)^2 = 100 p^2 + 180 p + 81 = 10 \left({10 p^2 + 18 p + 8}\right)+ 1$.
It is clear that in all cases the 10's digit is even.
So the square of every odd integer greater than $3$ always has at least one even digit.
Hence the result.
{{qed}}
\end{proof}
|
21757
|
\section{Squaring the Circle by Compass and Straightedge Construction is Impossible}
Tags: Classic Problems, Squaring the Circle, Euclidean Geometry, Solid Geometry
\begin{theorem}
There is no compass and straightedge construction to allow a square to be constructed whose area is equal to that of a given circle.
\end{theorem}
\begin{proof}
Squaring the Circle consists of constructing a line segment of length $\sqrt \pi$ of another.
From Constructible Length with Compass and Straightedge, any such line segment has a length which is an algebraic number of degree $2$.
But $\pi$ is transcendental.
Hence $\pi$ and therefore $\sqrt \pi$ is not such an algebraic number.
Therefore any attempt at such a construction will fail.
{{Qed}}
\end{proof}
|
21758
|
\section{Squeeze Theorem/Functions}
Tags: Named Theorems, Limits of Real Functions, Limits of Functions, Squeeze Theorem
\begin{theorem}
Let $a$ be a point on an open real interval $I$.
Let $f$, $g$ and $h$ be real functions defined at all points of $I$ except for possibly at point $a$.
Suppose that:
:$\forall x \ne a \in I: \map g x \le \map f x \le \map h x$
:$\ds \lim_{x \mathop \to a} \map g x = \lim_{x \mathop \to a} \map h x = L$
Then:
:$\ds \lim_{x \mathop \to a} \ \map f x = L$
\end{theorem}
\begin{proof}
We start by proving the special case where $\forall x: g \left({x}\right) = 0$ and $L=0$, in which case $\displaystyle \lim_{x \to a} \ h \left({x}\right) = 0$.
Let $\epsilon > 0$ be a positive real number.
Then by the definition of the limit of a function:
: $\exists \delta > 0: 0 < \left|{x - a}\right| < \delta \implies \left|{h \left({x}\right)}\right| < \epsilon$
Now:
: $\forall x \ne a: 0 = g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
so that:
:$\left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right|$
Thus:
: $0 < |x-a| < \delta \implies \left|{f \left({x}\right)}\right| \le \left|{h \left({x}\right)}\right| < \epsilon$
By the transitive property of $\le$, this proves that:
: $\displaystyle \lim_{x \to a} \ f \left({x}\right) = 0 = L$
We now move on to the general case, with $g \left({x}\right)$ and $L$ arbitrary.
For $x \ne a$, we have:
: $g \left({x}\right) \le f \left({x}\right) \le h \left({x}\right)$
By subtracting $g \left({x}\right)$ from all expressions, we have:
: $0 \le f \left({x}\right) - g \left({x}\right) \le h \left({x}\right) - g \left({x}\right)$
Since as $x \to a, h \left({x}\right) \to L$ and $g \left({x}\right) \to L$, we have:
: $h \left({x}\right) - g \left({x}\right) \to L - L = 0$
From the special case, we now have:
: $f \left({x}\right) - g \left({x}\right) \to 0$
We conclude that:
: $f \left({x}\right) = \left({f \left({x}\right) - g \left({x}\right)}\right) + g \left({x}\right) \to 0 + L = L$
{{qed}}
\end{proof}
|
21759
|
\section{Squeeze Theorem/Sequences/Complex Numbers}
Tags: Named Theorems, Limits of Sequences
\begin{theorem}
Let $\sequence {a_n}$ be a null sequence in $\R$, that is:
:$a_n \to 0$ as $n \to \infty$
Let $\sequence {z_n}$ be a sequence in $\C$.
Suppose $\sequence {a_n}$ dominates $\sequence {z_n}$.
That is:
: $\forall n \in \N: \cmod {z_n} \le a_n$
Then $\sequence {z_n}$ is a null sequence.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = \cmod {z_n}
| o = \le
| r = a_n
| c = {{Defof|Dominate (Analysis)}}
}}
{{eqn | q = \forall n \in \N
| l = a_n
| o = \le
| r = \size {a_n}
| c = Negative of Absolute Value
}}
{{eqn | q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
| lo= n > N \implies
| l = \size {a_n}
| o = <
| r = \epsilon
| c = {{Defof|Null Sequence|subdef = Real Numbers}}
}}
{{eqn | ll= \leadsto
| q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
| lo= n > N \implies
| l = \cmod {z_n}
| o = <
| r = \epsilon
| c = Extended Transitivity
}}
{{end-eqn}}
Thus $\sequence {z_n}$ is a null sequence.
{{qed}}
\end{proof}
|
21760
|
\section{Squeeze Theorem/Sequences/Linearly Ordered Space}
Tags: Order Topology, Limits of Sequences
\begin{theorem}
Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $\sequence {x_n}$, $\sequence {y_n}$, and $\sequence {z_n}$ be sequences in $S$.
Let $p \in S$.
Let $\sequence {x_n}$ and $\sequence {z_n}$ both converge to $p$.
Let $\forall n \in \N: x_n \le y_n \le z_n$.
Then $\sequence {y_n}$ converges to $p$.
\end{theorem}
\begin{proof}
Let $m \in S$ and $m < p$.
Then $\sequence {x_n}$ eventually succeeds $m$.
Thus by Extended Transitivity, $\sequence {y_n}$ eventually succeeds $m$.
A similar argument using $\sequence {z_n}$ proves the dual statement.
Thus $\sequence {y_n}$ is eventually in each ray containing $p$, so it converges to $p$.
{{qed}}
Category:Order Topology
Category:Limits of Sequences
\end{proof}
|
21761
|
\section{Squeeze Theorem/Sequences/Metric Spaces}
Tags: Metric Spaces, Limits of Sequences
\begin{theorem}
Let $M = \struct {S, d}$ be a metric space or pseudometric space.
Let $p \in S$.
Let $\sequence {r_n}$ be a null sequence in $\R$.
Let $\sequence {x_n}$ be a sequence in $S$ such that:
:$\forall n \in \N: \map d {p, x_n} \le r_n$.
Then $\sequence {x_n}$ converges to $p$.
\end{theorem}
\begin{proof}
{{begin-eqn}}
{{eqn | q = \forall n \in \N
| l = \map d {p, x_n}
| o = \le
| r = r_n
| c = {{hypothesis}}
}}
{{eqn | q = \forall n \in \N
| l = r_n
| o = \le
| r = \size {r_n}
| c = Negative of Absolute Value
}}
{{eqn | q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
| lo= n > N \implies
| l = \size {r_n}
| o = <
| r = \epsilon
| c = {{Defof|Null Sequence|subdef = Real Numbers}}
}}
{{eqn | ll= \leadsto
| q = \forall \epsilon \in \R_{>0}: \exists N \in \R_{>0}: \forall n \in \N
| lo= n > N \implies
| l = \map d {p, x_n}
| o = <
| r = \epsilon
| c = Extended Transitivity
}}
{{end-eqn}}
Thus $\sequence {x_n}$ converges to $p$.
{{qed}}
Category:Metric Spaces
Category:Limits of Sequences
\end{proof}
|
21762
|
\section{Squeeze Theorem/Sequences/Real Numbers}
Tags: Named Theorems, Limits of Sequences, Squeeze Theorem, Real Analysis
\begin{theorem}
Let $\sequence {x_n}$, $\sequence {y_n}$ and $\sequence {z_n}$ be sequences in $\R$.
Let $\sequence {y_n}$ and $\sequence {z_n}$ both be convergent to the following limit:
:$\ds \lim_{n \mathop \to \infty} y_n = l, \lim_{n \mathop \to \infty} z_n = l$
Suppose that:
:$\forall n \in \N: y_n \le x_n \le z_n$
Then:
:$x_n \to l$ as $n \to \infty$
that is:
:$\ds \lim_{n \mathop \to \infty} x_n = l$
Thus, if $\sequence {x_n}$ is always between two other sequences that both converge to the same limit, $\sequence {x_n} $ is said to be '''sandwiched''' or '''squeezed''' between those two sequences and itself must therefore converge to that same limit.
\end{theorem}
\begin{proof}
From Negative of Absolute Value: Corollary 1:
:$\size {x - l} < \epsilon \iff l - \epsilon < x < l + \epsilon$
Let $\epsilon > 0$.
We need to prove that:
:$\exists N: \forall n > N: \size {x_n - l} < \epsilon$
As $\ds \lim_{n \mathop \to \infty} y_n = l$ we know that:
:$\exists N_1: \forall n > N_1: \size {y_n - l} < \epsilon$
As $\ds \lim_{n \mathop \to \infty} z_n = l$ we know that:
:$\exists N_2: \forall n > N_2: \size {z_n - l} < \epsilon$
Let $N = \max \set {N_1, N_2}$.
Then if $n > N$, it follows that $n > N_1$ and $n > N_2$.
So:
:$\forall n > N: l - \epsilon < y_n < l + \epsilon$
:$\forall n > N: l - \epsilon < z_n < l + \epsilon$
But:
:$\forall n \in \N: y_n \le x_n \le z_n$
So:
:$\forall n > N: l - \epsilon < y_n \le x_n \le z_n < l + \epsilon$
and so:
:$\forall n > N: l - \epsilon < x_n < l + \epsilon$
So:
:$\forall n > N: \size {x_n - l} < \epsilon$
Hence the result.
{{qed}}
\end{proof}
|
21763
|
\section{Squeeze Theorem for Absolutely Convergent Series}
Tags: Series
\begin{theorem}
Let $\ds \sum \size {a_n}$ be an absolutely convergent series in $\R$.
Suppose that:
:$\ds -\sum \size {a_n} = \sum \size {a_n}$
Then $\ds \sum a_n$ equals the above two series.
\end{theorem}
\begin{proof}
From Absolutely Convergent Series is Convergent, the convergence of:
{{refactor|We really want to be linking to Absolutely Convergent Series is Convergent/Real Numbers here, because the student who is working through basic undergraduate analysis won't know (or even care) what a Banach space is. Consequently, Absolutely Convergent Series is Convergent/Real Numbers can be given a proof which basically goes "the real numbers form a Banach space, the result follows from Absolutely Convergent Series is Convergent iff Normed Vector Space is Banach" which gives us an in from both undergrad analysis and this more rarefied topological / functional-analytical level.}}
:$\ds \sum_{n \mathop = 1}^\infty \size {a_n}$
implies that of:
:$\ds \sum_{n \mathop = 1}^\infty a_n$
By Negative of Absolute Value:
:$\ds -\size {\sum_{n \mathop = 1}^j a_n} \le \sum_{n \mathop = 1}^j a_n \le \size {\sum_{n \mathop = 1}^j a_n}$
By repeated application of Triangle Inequality:
:$\ds -\sum_{n \mathop = 1}^j \size {a_n} \le \sum_{n \mathop = 1}^j a_n \le \sum_{n \mathop = 1}^j \size {a_n}$
By hypothesis, the leftmost and rightmost terms converge as $j \to +\infty$.
Hence the result, from Squeeze Theorem for Real Sequences.
{{qed}}
Category:Series
\end{proof}
|
21764
|
\section{Squeeze Theorem for Filter Bases}
Tags: Filter Theory, Order Topology, Convergence
\begin{theorem}
Let $\struct {S, \le, \tau}$ be a linearly ordered space.
Let $F_1$, $F_2$, and $F_3$ be filter bases in $S$.
Let:
:$\forall T \in F_1: \exists M \in F_2: \forall x \in M: \exists y \in T: y \le x$
That is:
:for each $T \in F_1$, $F_2$ has an element $M$ such that all elements of $M$ succeed some element of $T$.
Similarly, let:
:$\forall U \in F_3: \exists N \in F_2: \forall x \in N: \exists y \in U: x \le y$
That is:
:for each $U \in F_3$, $F_2$ has an element $N$ such that all elements of $N$ precede some element of $U$.
Let $F_1$ and $F_3$ each converge to a point $p \in S$.
Then $F_2$ converges to $p$.
\end{theorem}
\begin{proof}
Let $q \in S$ such that $q < p$.
We will show that $F_2$ has an element which is a subset of $q^\ge$.
{{explain|$q^\ge$ is the upper closure in what set?}}
Since $F_1$ converges to $p$, it has an element:
:$A \subseteq q^\ge$.
Thus there is an element $k$ in $A$ and an element $M$ in $F_2$ such that all elements of $M$ succeed $k$.
Then by Extended Transitivity:
:$M \subseteq q^\ge$
A similar argument using $F_3$ proves the dual statement.
Thus $F_2$ converges to $p$.
{{explain}}
{{qed}}
Category:Filter Theory
Category:Order Topology
Category:Convergence
\end{proof}
|
21765
|
\section{Stabilizer in Group of Transformations}
Tags: Symmetric Groups, Stabilizers, Group Actions, Symmetric Group
\begin{theorem}
Let $X$ be any set with $n$ elements (where $n \in \Z_{>0}$).
Consider the symmetric group on $n$ letters $S_n$ as a group of transformations on $X$.
Let $x \in X$.
Then the stabilizer of $x$ is isomorphic to $S_{n - 1}$.
\end{theorem}
\begin{proof}
Consider the initial segment of natural numbers $\N_n = \set {1, 2, \ldots, n}$.
By the definition of cardinality, $H$ is equivalent to $\N_n$.
{{WLOG}} we can consider $S_n$ acting directly on $\N_n$.
The stabilizer of $n$ in $\N_n$ is all the permutations of $S_n$ which fix $n$, which is clearly $S_{n - 1}$.
A permutation can be applied to $\N_n$ so that $i \to n$ for any $i$.
Thus one can build an isomorphism to show the result for a general $i$.
{{qed}}
\end{proof}
|
21766
|
\section{Stabilizer is Subgroup}
Tags: Group Actions, Stabilizer is Subgroup, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group which acts on a set $X$.
Let $\Stab x$ be the stabilizer of $x$ by $G$.
Then for each $x \in X$, $\Stab x$ is a subgroup of $G$.
\end{theorem}
\begin{proof}
From the {{GroupActionAxiom|2}}:
:$e * x = x \implies e \in \Stab x$
and so $\Stab x$ cannot be empty.
Let $g, h \in \Stab x$.
{{begin-eqn}}
{{eqn | l = g, h
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadsto
| l = g * x
| r = x
| c = {{Defof|Stabilizer}} of $x$ by $G$
}}
{{eqn | lo= \land
| l = h * x
| r = x
| c = {{Defof|Stabilizer}} of $x$ by $G$
}}
{{eqn | ll= \leadsto
| l = g * \paren {h * x}
| r = x
| c =
}}
{{eqn | ll= \leadsto
| l = \paren {g \circ h} * x
| r = x
| c = {{GroupActionAxiom|1}}
}}
{{eqn | ll= \leadsto
| l = g \circ h
| o = \in
| r = \Stab x
| c = {{Defof|Stabilizer}} of $x$ by $G$
}}
{{end-eqn}}
Let $g \in \Stab x$.
Then:
:$x = \paren {g^{-1} \circ g} * x = g^{-1} * \paren {g * x} = g^{-1} * x$
Hence $g^{-1} \in \Stab x$.
Thus the conditions for the Two-Step Subgroup Test are fulfilled, and $\Stab x \le G$.
{{qed}}
\end{proof}
|
21767
|
\section{Stabilizer of Cartesian Product of Group Actions}
Tags: Group Actions, Cartesian Product, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group.
Let $S$ and $T$ be sets.
Let $*_S: G \times S \to S$ and $*_T: G \times T \to T$ be group actions.
Let the group action $*: G \times \paren {S \times T} \to S \times T$ be defined as:
:$\forall \tuple {g, \tuple {s, t} } \in G \times \paren {S \times T}: g * \tuple {s, t} = \tuple {g *_S s, g *_T t}$
Then the stabilizer of $\tuple {s, t} \in S \times T$ is given by:
:$\Stab {s, t} = \Stab s \cap \Stab t$
where $\Stab s$ and $\Stab t$ are the stabilizers of $s$ and $t$ under $*_S$ and $*_T$ respectively.
\end{theorem}
\begin{proof}
By definition, the stabilizer of an element $x$ of $S$ is defined as:
:$\Stab x := \set {g \in G: g * x = x}$
where $*$ denotes the group action.
So:
{{begin-eqn}}
{{eqn | l = \Stab {s, t}
| r = \set {g \in G: g * \tuple {s, t} = \tuple {s, t} }
| c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {g \in G: \tuple {g *_S s, g *_T t} = \tuple {s, t} }
| c = Definition of $*$
}}
{{eqn | r = \set {g \in G: g *_S s = s \land g *_T t = t}
| c = {{Defof|Cartesian Product}}
}}
{{eqn | r = \set {g \in G: g \in \Stab s \land g \in \Stab t}
| c = {{Defof|Stabilizer}}
}}
{{eqn | r = \Stab s \cap \Stab t
| c = {{Defof|Set Intersection}}
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21768
|
\section{Stabilizer of Conjugacy Action on Subgroup is Normalizer}
Tags: Conjugacy, Conjugacy Action, Group Actions, Normalizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $X$ be the set of all subgroups of $G$.
Let $*$ be the conjugacy action on $H$:
:$\forall g \in G, H \in X: g * H = g \circ H \circ g^{-1}$
Then the stabilizer of $H$ in $\powerset G$ is given by:
:$\Stab H = \map {N_G} H$
where $\map {N_G} H$ is the normalizer of $H$ in $G$.
\end{theorem}
\begin{proof}
We have that:
:$\Stab H = \set {g \in G: g \circ H \circ g^{-1} = H}$
which is precisely how the normalizer is defined.
{{Qed}}
\end{proof}
|
21769
|
\section{Stabilizer of Coset Action on Set of Subgroups}
Tags: Subset Product Action, Coset Action, Group Actions, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $\powerset G$ denote the power set of $G$.
Let $\HH \subseteq \powerset G$ denote the set of subgroups of $G$.
Let $*$ be the subset product action on $\HH \subseteq \powerset G$ defined as:
:$\forall g \in G: \forall H \in \HH: g * H = g \circ H$
where $g \circ H$ is the (left) coset of $g$ by $H$.
Then the stabilizer of $H$ in $\powerset G$ is $H$ itself:
:$\Stab H = H$
\end{theorem}
\begin{proof}
From the definition of Stabilizer of Subset Product Action on Power Set:
:$\Stab H = H = \set {g \in G: g * H = H}$
The result follows from Left Coset Equals Subgroup iff Element in Subgroup.
{{qed}}
\end{proof}
|
21770
|
\section{Stabilizer of Coset under Group Action on Coset Space}
Tags: Group Action on Coset Space, Stabilizers
\begin{theorem}
Let $G$ be a group whose identity is $e$.
Let $H$ be a subgroup of $G$.
Let $*: G \times G / H \to G / H$ be the action on the (left) coset space:
:$\forall g \in G, \forall g' H \in G / H: g * \paren {g' H} := \paren {g g'} H$
Then the stabilizer of $a H$ under $*$ is given by:
:$\Stab {a H} = a H a^{-1}$
\end{theorem}
\begin{proof}
It is established in Action of Group on Coset Space is Group Action that $*$ is a group action.
Then:
{{begin-eqn}}
{{eqn | l = \Stab {a H}
| r = \set {g \in G: g * a H = a H}
| c = {{Defof|Stabilizer}}
}}
{{eqn | r = \set {g \in G: \paren {g a} H = a H}
| c =
}}
{{eqn | r = \set {g \in G: g H = a H a^{-1} }
| c =
}}
{{eqn | r = a H a^{-1}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
21771
|
\section{Stabilizer of Element of Group Acting on Itself is Trivial}
Tags: Group Actions, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the group action of $\struct {G, \circ}$ on itself by the rule:
:$\forall g, h \in G: g * h = g \circ h$
Then the stabilizer of an element $x \in G$ is given by:
:$\Stab x = \set e$
\end{theorem}
\begin{proof}
Let $g \in \Stab x$.
Then:
{{begin-eqn}}
{{eqn | l = g * x
| r = x
| c = {{Defof|Stabilizer}}
}}
{{eqn | ll= \leadsto
| l = g \circ x
| r = x
| c = {{Defof|Group Action}} (this particular one)
}}
{{eqn | ll= \leadsto
| l = g
| r = e
| c = {{Defof|Identity Element}}
}}
{{end-eqn}}
Hence the result, by definition of trivial subgroup.
{{qed}}
\end{proof}
|
21772
|
\section{Stabilizer of Element under Conjugacy Action is Centralizer}
Tags: Centralizers, Conjugacy Action, Group Actions, Stabilizers
\begin{theorem}
Let $\struct {G, \circ}$ be a group whose identity is $e$.
Let $*$ be the conjugacy action on $G$ defined by the rule:
:$\forall g, h \in G: g * h = g \circ h \circ g^{-1}$
Let $x \in G$.
Then the stabilizer of $x$ under this conjugacy action is:
:$\Stab x = \map {C_G} x$
where $\map {C_G} x$ is the centralizer of $x$ in $G$.
\end{theorem}
\begin{proof}
From the definition of centralizer:
:$\map {C_G} x = \set {g \in G: g \circ x = x \circ g}$
Then:
{{begin-eqn}}
{{eqn | l = z
| o = \in
| r = \Stab x
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = \in
| r = \set {g \in G: g \circ x \circ g^{-1} = x}
| c =
}}
{{eqn | ll= \leadstoandfrom
| l = z
| o = \in
| r = \set {g \in G: g \circ x = x \circ g}
| c =
}}
{{end-eqn}}
{{qed}}
\end{proof}
|
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